FSP Training 04

Preparation for Mock Exam 05

Instructions

This training page covers all of Section 04 and prepares you for Mock Exam 05. Set A trains setting up: you translate a text into mathematics and check your translation. Set B is a practice mock in the format of Mock Exam 05: 90 minutes, 50 points, and all 3 problems must be completed. Write down every step as you would in the exam. The points show what each part is worth.

In the real FSP, decimals are written with a comma (1,25 means 1.25).

Set A: Set Up & Check (40 minutes)

From Text to Mathematics (xx)

Translate each statement into mathematics. Then answer the check question: it tells you whether your translation works.

1. An FSP task shows the graph of a polynomial function \(f\) of third degree. The graph crosses the \(x\)-axis at \(x = -2\), touches it at \(x = 3\) and crosses the \(y\)-axis at \((0, -9)\). The task says: “Determine the function term of \(f\) in factored form.” Give \(f(x)\).

   Check: The graph also passes through the point \((1, -6)\). Does your term give this value?

2. An FSP task says: “Show that \(x = -3\) is a zero of \(f(x) = x^3 - 13x - 12\). Determine all zeros of \(f\) using polynomial long division.” Write down the division you have to carry out, with the dividend and the divisor.

   Hint: \(f\) has no \(x^2\)-term. Write it as \(0x^2\) in the long division, so that every power has its own column.

   Check: Which remainder must your division give? Carry out the division and check it by multiplying back.

3. An FSP task says: “Determine the domain of \(g(x) = \frac{x^2 + x - 6}{x^2 - 9}\).” A student first cancels, gets \(g(x) = \frac{x - 2}{x - 3}\) and writes the domain \(\mathbb{R} \setminus \{3\}\). Give the correct domain.

   Check: Is \(g(-3)\) defined? What does the graph of \(g\) look like at \(x = -3\)?

4. An FSP task says: “Decide whether the graph of \(h\) lies above or below its oblique asymptote for large positive \(x\).” Polynomial long division has already given \(h(x) = \frac{-x^2 + 4x + 1}{x - 1} = -x + 3 + \frac{4}{x - 1}\). Write the question as an inequality and decide.

   Check: Compute \(h(9)\) and \(h(-7)\) and compare them with the values of the asymptote at \(x = 9\) and \(x = -7\).

5. “A regional newspaper has 42,000 print subscribers today and loses 2 % of them every year. The publisher asks: ‘After how many whole years will we have fewer than 35,000 print subscribers for the first time?’” Set up the inequality with \(t\) in years and solve it.

   Hint: The larger of two positive numbers has the larger logarithm, so taking the logarithm of both sides keeps the inequality sign. The logarithm of a number between 0 and 1 is negative.

   Check: Compute the number of print subscribers after your number of years and one year earlier.

6. “A tea merchant has opened an online shop. Its weekly orders are modelled by \(S(t) = L - a \cdot b^t\) with \(t\) in weeks, \(a > 0\) and \(0 < b < 1\). In the first week (\(t = 0\)) there are 200 orders. The orders approach 500 per week, and the gap to 500 shrinks by 20 % every week.” Give \(L\), \(a\) and \(b\).

   Check: Compute \(S(1)\). Is the gap to 500 after one week 20 % smaller than at the start?

7. “A supermarket that is open 24 hours a day counts its customers per hour. The number is highest at 19:00 with 350 customers and lowest at 7:00 with 90 customers, and the pattern repeats every day.” Give the model \(N(t) = A\cos(B(t - C)) + D\) with \(t\) in hours after midnight.

   Check: Compute \(N(7)\) without a calculator and \(N(16)\) with the calculator. Which indicator must show? A classmate gets about 350 customers at 16:00. What went wrong?

8. “A packaging machine moves a conveyor belt over a drum with a radius of 0.3 m. In each work step, the drum turns by the angle \(\frac{5\pi}{3}\).” Give the distance the belt moves in one work step.

   Check: Which fraction of a full turn is \(\frac{5\pi}{3}\)? Does your distance match this fraction of the circumference?

9. An FSP task says: “Give a function \(f_1(t)\) which is the mirror image of \(f(t) = 8 \cdot e^{0.25t}\) at the \(y\)-axis and give a reason for your choice.” Give \(f_1(t)\) and the reason.

   Check: Compute \(f(4)\) and \(f_1(-4)\). Which result do you expect?

CautionSolution

1. Where the graph touches the \(x\)-axis, the zero has an even multiplicity. Where it crosses, the multiplicity is odd. With degree 3, only \(2 + 1 = 3\) fits: \(x = 3\) is a double zero and \(x = -2\) a simple zero. Ansatz with the unknown factor \(a\), then insert the \(y\)-intercept \((0, -9)\):

   \[f(x) = a(x + 2)(x - 3)^2, \qquad f(0) = a \cdot 2 \cdot 9 = 18a = -9 \iff a = -0.5\]

   \[f(x) = -0.5(x + 2)(x - 3)^2\]

   Check: yes, \(f(1) = -0.5 \cdot 3 \cdot 4 = -6\).

   Typical mistake: \(f(x) = (x + 2)(x - 3)^2\) without the factor \(a\). Then \(f(0) = 18\) instead of \(-9\). A second mistake is the square at the wrong zero, \(a(x + 2)^2(x - 3)\): this graph would touch the \(x\)-axis at \(x = -2\).

2. \(f(-3) = -27 + 39 - 12 = 0\) ✓. The zero \(x = -3\) gives the factor \(x - (-3) = x + 3\). \(f\) has no \(x^2\)-term, so write it as \(0x^2\): every power needs its own column.

   \[(x^3 + 0x^2 - 13x - 12) : (x + 3)\]

   Check: the remainder must be 0, because \(-3\) is a zero.

   \[\begin{array}{l} (x^3 + 0x^2 - 13x - 12) : (x + 3) = x^2 - 3x - 4 \\ \underline{-(x^3 + 3x^2)} \\ \phantom{(x^3}\;\; -3x^2 - 13x \\ \phantom{(x^3}\;\; \underline{-(-3x^2 - 9x)} \\ \phantom{(x^3 - 3x^2}\;\; -4x - 12 \\ \phantom{(x^3 - 3x^2}\;\; \underline{-(-4x - 12)} \\ \phantom{(x^3 - 3x^2 - 4x - 1}\;\; 0 \end{array}\]

   Multiplying back: \((x + 3)(x^2 - 3x - 4) = x^3 - 3x^2 - 4x + 3x^2 - 9x - 12 = x^3 - 13x - 12\) ✓. With \(x^2 - 3x - 4 = (x - 4)(x + 1)\), the zeros are \(x = -3\), \(x = -1\) and \(x = 4\).

   Typical mistake: dividing by \((x - 3)\). Then the division does not come out even (the remainder is \(-24\)), because \(3\) is not a zero: \(f(3) = -24\). Without \(0x^2\), the terms \(-13x\) and \(-12\) slip into the wrong columns.

3. The domain comes from the term in the task, before cancelling. The denominator must not be zero:

   \[x^2 - 9 = (x + 3)(x - 3) = 0 \iff x = -3 \text{ or } x = 3, \qquad D_g = \mathbb{R} \setminus \{-3, 3\}\]

   With \(x^2 + x - 6 = (x + 3)(x - 2)\), the simplified term \(g(x) = \frac{x - 2}{x - 3}\) holds only for \(x \neq -3\) and \(x \neq 3\).

   Check: no, \(g(-3)\) is not defined, because the denominator in the task is zero there. The factor \(x + 3\) cancels, so the graph has a hole at \(x = -3\), not an asymptote. Its \(y\)-value comes from the simplified term, \(\frac{-3 - 2}{-3 - 3} = \frac{5}{6}\): the hole is at \(\left(-3, \frac{5}{6}\right)\). At \(x = 3\), the factor \(x - 3\) stays in the denominator: vertical asymptote \(x = 3\).

   Typical mistake: taking the domain from the simplified term. Cancelling does not make \(g\) defined at \(x = -3\).

4. The oblique asymptote is \(y = -x + 3\). The graph lies above it if

   \[h(x) > -x + 3 \iff \frac{4}{x - 1} > 0\]

   For \(x > 1\), the denominator \(x - 1\) is positive, so \(\frac{4}{x - 1} > 0\): for large positive \(x\), the graph lies above the asymptote.

   Check: \(h(9) = -6 + \frac{4}{8} = -5.5\), and the asymptote gives \(-9 + 3 = -6\): above ✓. \(h(-7) = 10 + \frac{4}{-8} = 9.5\), and the asymptote gives \(10\): here the graph lies below. For \(x < 1\), the denominator is negative, and so is the remainder term.

   Typical mistake: “the remainder 4 is positive, so the graph always lies above the asymptote”. The sign of \(\frac{4}{x - 1}\) also depends on the denominator: for large negative \(x\), the graph lies below.

5. Losing 2 % every year multiplies the number by \(0.98\): \(S(t) = 42000 \cdot 0.98^t\). Condition:

   \[42000 \cdot 0.98^t < 35000 \iff 0.98^t < \frac{5}{6}\]

   Take the logarithm of both sides (the sign stays: the larger of two positive numbers has the larger logarithm) and use the power rule: \(t \cdot \ln(0.98) < \ln\left(\frac{5}{6}\right)\). Since \(\ln(0.98) < 0\), dividing by it flips the sign:

   \[t > \frac{\ln\left(\frac{5}{6}\right)}{\ln(0.98)} \approx 9.02\]

   The first whole number above 9.02 is 10: after 10 years, there are fewer than 35,000 print subscribers for the first time.

   Check: \(S(9) = 42000 \cdot 0.98^9 \approx 35{,}017\) (not yet below 35,000) and \(S(10) = 42000 \cdot 0.98^{10} \approx 34{,}317\) ✓.

   Typical mistake: keeping the sign when dividing by \(\ln(0.98)\). Then \(t < 9.02\), i.e. “within 9 years”, but \(S(9)\) is still above 35,000. Rounding 9.02 down to 9 gives the same wrong answer.

6. For \(t \to \infty\), \(b^t \to 0\) (since \(0 < b < 1\)), so \(S(t)\) approaches \(L\): \(L = 500\). The gap to 500 is \(500 - S(t) = a \cdot b^t\).

   • Start: \(S(0) = 500 - a = 200 \iff a = 300\), the gap at the start
   • The gap shrinks by 20 % every week, so it is multiplied by \(b = 0.8\)

   \[S(t) = 500 - 300 \cdot 0.8^t\]

   Check: \(S(1) = 500 - 240 = 260\). The gap is now \(240 = 0.8 \cdot 300\), i.e. 20 % smaller than 300 ✓.

   Typical mistake: \(b = 0.2\). Then the gap would shrink by 80 % per week, and \(S(1) = 440\). A second mistake is \(a = 200\), the start value: then \(S(0) = 300\). The factor \(a\) is the gap at the start, not the start value.

7. Amplitude \(A = \frac{350 - 90}{2} = 130\), midline \(D = \frac{350 + 90}{2} = 220\), period 24 hours, so \(B = \frac{2\pi}{24} = \frac{\pi}{12}\). A cosine starts at its maximum. The maximum is at 19:00, so the cosine is shifted 19 hours to the right: \(C = 19\).

   \[N(t) = 220 + 130\cos\left(\frac{\pi}{12}(t - 19)\right)\]

   Check: \(N(7) = 220 + 130\cos(-\pi) = 220 - 130 = 90\) ✓ (the angles \(-\pi\) and \(\pi\) lead to the same point \((-1, 0)\) on the unit circle). \(N(16) = 220 + 130\cos\left(-\frac{\pi}{4}\right) \approx 311.92\), i.e. about 312 customers. The angle \(-\frac{\pi}{4} \approx -0.7854\) is in radians, so the indicator R must show. The classmate’s calculator showed D: it took \(-0.7854\) as degrees, and \(\cos(-0.7854°) \approx 0.9999\) gives almost the maximum of 350.

   Typical mistake: \(C = -19\), i.e. \(\cos\left(\frac{\pi}{12}(t + 19)\right)\). This curve has its maximum at \(t = -19 + 24 = 5\), i.e. at 5:00. Minus in the bracket shifts to the right.

8. The arc length is \(s = r\theta\) with the angle \(\theta\) in radians:

   \[s = 0.3 \cdot \frac{5\pi}{3} = 0.5\pi \approx 1.57 \text{ m}\]

   Check: \(\frac{5\pi}{3} : 2\pi = \frac{5}{6}\) of a full turn. The circumference is \(2\pi \cdot 0.3 = 0.6\pi\) m, and \(\frac{5}{6} \cdot 0.6\pi = 0.5\pi\) ✓.

   Typical mistake: using degrees. \(\frac{5\pi}{3}\) is \(300°\), and \(0.3 \cdot 300 = 90\) m would be far too long. \(s = r\theta\) only holds with the angle in radians.

9. Replace \(t\) by \(-t\):

   \[f_1(t) = f(-t) = 8 \cdot e^{-0.25t}\]

   Reason: replacing \(t\) by \(-t\) maps each point \((a, f(a))\) of the graph of \(f\) to the point \((-a, f(a))\). Every point keeps its height and moves to the other side of the \(y\)-axis at the same distance. The growth becomes a decay with the same rate.

   Check: both values should be equal, because \((-4, f(4))\) is the mirror image of \((4, f(4))\). Indeed \(f(4) = 8e^{1} \approx 21.75\) and \(f_1(-4) = 8e^{1} \approx 21.75\) ✓.

   Typical mistake: \(f_1(t) = -8 \cdot e^{0.25t}\). This is the mirror image at the \(t\)-axis, and all its values are negative. The mirror image at the \(y\)-axis changes the input, the mirror image at the \(t\)-axis changes the output.

One Paragraph, All Conditions (xx)

A company builds espresso machines for offices and sells them directly to the offices. It models its monthly total costs by a polynomial of third degree. Its fixed costs are 90,000 EUR per month. The variable costs (they grow with the output) of 200 machines are 52,000 EUR, and 400 machines cause total costs of 162,000 EUR. With 500 machines, the average cost is 380 EUR per machine. The offices buy no machine at a price of 1,500 EUR or more (maximum price), and at a price of 0 EUR they would take 3,000 machines per month (market saturation). In between, the demand is linear. The company can build at most 900 machines per month. The owner wants to know the cost function, the profit function, and the profit and the price for the planned output of 600 machines per month.

(QU = quantity units, MU = monetary units. 1 QU = 100 machines, 1 MU = 1,000 EUR, so a price of \(p\) MU per QU is a price of \(10p\) EUR per machine)

List every condition and every function in this paragraph as a mathematical statement, with \(x\) in QU and all money in MU. Do not solve anything.

Check: An assistant writes the condition for 500 machines as \(C(5) = 380\). Name the two mistakes.

CautionSolution

1. Ansatz (polynomial of third degree): \(C(x) = ax^3 + bx^2 + cx + d\)
2. Fixed costs: 90,000 EUR \(= 90\) MU, so \(C(0) = d = 90\)
3. 200 machines \(= 2\) QU: the variable costs are \(C(2) - 90 = 52\), so \(C(2) = 142\)
4. 400 machines: \(C(4) = 162\)
5. 500 machines: 380 EUR per machine \(= 38{,}000\) EUR per QU \(= 38\) MU per QU. Average cost \(\frac{C(5)}{5} = 38\), so \(C(5) = 190\)
6. Price-demand function: the maximum price 1,500 EUR per machine is 150 MU per QU, the market saturation 3,000 machines is 30 QU. From \(p(0) = 150\) and \(p(30) = 0\): \(p(x) = 150 - 5x\)
7. Revenue: \(R(x) = x \cdot p(x) = 150x - 5x^2\)
8. Profit: \(P(x) = R(x) - C(x)\)
9. Domain: \(0 \leq x \leq 9\) (the capacity of 900 machines lies below the market saturation)
10. Planned output 600 machines \(= 6\) QU: the profit \(P(6)\) in MU (times 1,000 for EUR) and the price \(p(6)\) in MU per QU (times 10 for EUR per machine)

Check: 380 is the average cost, not the total costs, and it is in EUR per machine, not in MU. The correct condition is \(\frac{C(5)}{5} = 38\), i.e. \(C(5) = 190\): 190,000 EUR for 500 machines, and \(\frac{190{,}000}{500} = 380\) EUR per machine ✓.

Typical mistake: \(C(2) = 52\). The 52,000 EUR are only the variable costs. The total costs of 200 machines also contain the fixed costs: \(C(2) = 52 + 90 = 142\).

FSP Words So Far

The FSP problems use these words from Sections 01 to 04. Learn what each one means and how to write it in mathematics.

Word	Meaning	Mathematics
quantity unit (QU)	a fixed number of items, e.g. 1 QU = 100 items	the output \(x\) is counted in QU
monetary unit (MU)	a fixed amount of money, e.g. 1 MU = 100 EUR	results in MU, convert to EUR at the end
fixed costs	costs that arise even if nothing is produced	the constant term of \(C(x)\)
variable costs	costs that grow with the output	output times costs per QU
total costs	fixed costs + variable costs	\(C(x)\)
revenue	quantity times price	\(R(x) = x \cdot \text{price}\)
profit	revenue − total costs	\(P(x) = R(x) - C(x)\)
break-even point	the output at which the profit changes from negative (a loss) to positive	\(P(x) = 0\), the smaller zero
profit limit (upper break-even point)	the larger output with zero profit: above it, the profit turns negative again	\(P(x) = 0\), the larger zero
profit zone	all outputs with a positive profit	\(P(x) > 0\): from the break-even point to the profit limit (or the capacity)
capacity	the maximum possible output	an upper bound: \(x \leq\) capacity
growth factor	the factor by which a quantity is multiplied each period	\(q\) in (starting value) \(\cdot\, q^t\)
growth rate	the change per period in percent	\(q = 1 + r\) with \(r\) as a decimal, a loss of 4 % gives \(q = 0.96\)
doubling time	the time after which a quantity has doubled	\(q^T = 2\), continuous: \(e^{kT} = 2\), so \(T = \frac{\ln 2}{k}\)
half-life (halving time)	the time after which a quantity has halved	\(q^T = 0.5\), continuous: \(e^{-kT} = 0.5\), so \(T = \frac{\ln 2}{k}\)
“deviates from \(a\) by at most \(d\)” (tolerance)	the distance from \(a\) is at most \(d\)	\(\lvert x - a \rvert \leq d \iff a - d \leq x \leq a + d\)
“share of \(A\) among \(B\)”	the part of \(B\) that also belongs to \(A\): the base is \(B\)	\(\dfrac{\text{number in } A \cap B}{\text{number in } B}\)
“neither \(A\) nor \(B\)”	not in \(A\) and not in \(B\), i.e. outside both circles	\(\overline{A \cup B}\)
contingency table	a table of counts for two properties: the inner fields are the regions of the Venn diagram, the margins are the totals	inner fields \(A \cap B\), \(A \cap \overline{B}\), \(\overline{A} \cap B\), \(\overline{A} \cap \overline{B}\)
gross price, net price	gross price = net price + VAT	with 19 % VAT: gross \(= 1.19 \cdot\) net
system of linear equations	several linear equations that must hold at the same time	solve by substitution or elimination, check the solution in every equation
no solution, infinitely many solutions	the conditions contradict each other, or one condition only repeats another	elimination leaves \(0 = d\) with \(d \neq 0\), or \(0 = 0\)
demand, supply	the quantity buyers want to buy and the quantity sellers offer at a given price	as terms in the price \(p\), e.g. \(Q_d(p)\), \(Q_s(p)\), or as prices \(p_D(x)\), \(p_S(x)\) for the quantity \(x\)
market equilibrium	the quantity demanded equals the quantity supplied	demand = supply: \(Q_d(p) = Q_s(p)\), or \(p_D(x) = p_S(x)\)
price-demand function	gives the price at which \(x\) QU can be sold	\(p(x)\), revenue \(R(x) = x \cdot p(x)\)
average cost	total costs divided by the number of units	\(\dfrac{C(x)}{x}\) with \(x \neq 0\)
zero form	all terms on one side, 0 on the other side	\(ax^2 + bx + c = 0\), then factor or use the quadratic formula
discriminant	decides how many real solutions a quadratic equation has	\(\Delta = b^2 - 4ac\): \(> 0\) two, \(= 0\) one, \(< 0\) none
biquadratic equation	an equation with only \(x^4\), \(x^2\) and a constant	substitute \(u = x^2\) in \(ax^4 + bx^2 + c\), solve for \(u\), then \(x = \pm\sqrt{u}\) for \(u \geq 0\)
domain restrictions (of an equation)	the conditions the unknown must meet so that every term is defined	denominators \(\neq 0\), under a root \(\geq 0\), inside a logarithm \(> 0\)
extraneous solution	a candidate that solves the squared equation but not the original one	check every candidate in the original equation
function	assigns to every input exactly one output	\(f(x)\), its graph passes the vertical line test
domain (of a function)	all inputs for which the function is defined, in a context only the sensible ones	e.g. \([0; \text{capacity}]\)
range	all outputs that the function takes on its domain	if the function grows or falls: between the values at the two ends, for a parabola also check the vertex
maximum price	the price at which nobody buys any more	\(Q_d(p) = 0\), or \(c\) in \(p = c - dQ\)
market saturation	the quantity demanded at a price of 0	\(a\) in \(Q_d = a - bp\), or the zero of \(p(x)\)
shortage, surplus	at a price below the equilibrium price, demand exceeds supply (shortage), above it, supply exceeds demand (surplus)	\(Q_d(p) > Q_s(p)\), or \(Q_d(p) < Q_s(p)\)
contribution margin	price minus variable costs per unit: the part of the price that covers the fixed costs	break-even quantity \(= \dfrac{\text{fixed costs}}{\text{contribution margin}}\)
vertex	the highest or lowest point of a parabola	\(x_v = -\dfrac{b}{2a}\), or \((h, k)\) in \(a(x - h)^2 + k\)
profit-maximizing output	the output with the highest profit	\(x_v\) of \(P(x)\), a maximum since \(a < 0\), and the price \(p(x_v)\). If the capacity lies below \(x_v\): the capacity and its price
shift, stretch, compression	moving or scaling a graph	\(a \cdot f(b(x - h)) + k\): first \(b\), then \(h\), then \(a\), then \(k\)
mirror image at the \(y\)-axis	the graph flipped from right to left	\(f(-x)\)
mirror image at the \(x\)-axis	the graph flipped upside down	\(-f(x)\)
composite function	first the inner function, then the outer function	\((f \circ g)(x) = f(g(x))\)
inverse function	reverses a function: gives the input that belongs to an output	solve \(y = f(x)\) for \(x\), e.g. \(x(p)\) from \(p(x)\)
polynomial of third degree	a function with powers of \(x\) up to \(x^3\)	\(ax^3 + bx^2 + cx + d\) with \(a \neq 0\): four unknowns, so four conditions
degree, leading coefficient	the highest power of \(x\), and the number in front of it	together they decide the end behavior
zero of multiplicity \(m\)	the factor \((x - c)\) appears \(m\) times in the factored form	odd \(m\): the graph crosses the \(x\)-axis at \(c\), even \(m\): it touches it
polynomial long division	divides a polynomial by \((x - c)\) once \(c\) is a known zero	\(f(x) = (x - c) \cdot q(x)\), the remainder must be 0
asymptotic behavior (end behavior)	what happens to \(f(x)\) for \(x \to \infty\) and \(x \to -\infty\)	polynomial: degree and sign of the leading coefficient. Rational function: its asymptotes
intercepts	the points where the graph meets the axes	\(y\)-intercept \((0, f(0))\), \(x\)-intercepts from \(f(x) = 0\)
rational function	a quotient of two polynomials	\(\dfrac{p(x)}{q(x)}\) with the domain \(q(x) \neq 0\)
vertical asymptote	a vertical line near which the values grow without bound	a zero of the denominator that does not cancel
hole	a single missing point of the graph	a factor that cancels, its \(y\)-value comes from the simplified term
horizontal asymptote	a horizontal line that the graph approaches for \(x \to \pm\infty\)	\(y = c\) with \(f(x) \to c\)
oblique asymptote	a slanted line that the graph approaches for \(x \to \pm\infty\)	the linear part of the long division, if the numerator’s degree is one higher
grows less than proportionally	grows, but twice the input gives less than twice the output	\(a \cdot x^b\) with \(0 < b < 1\)
continuous growth rate	the rate \(k\) of a model with the base \(e\)	\(N_0 \cdot e^{kt}\), growth factor per period \(e^k\)
saturation level	a level that a quantity approaches but never exceeds	a horizontal asymptote \(y = L\), e.g. of \(L - a \cdot e^{-kt}\) or \(\dfrac{L}{1 + A e^{-kt}}\)
behavior for \(t \to \infty\)	the value a quantity approaches in the long run	\(N(t) \to c\) for \(t \to \infty\), often with \(e^{-kt} \to 0\)
radian	an angle measure in which a full turn is \(2\pi\)	\(180° = \pi\), arc length \(s = r\theta\)
amplitude	half the distance between the highest and the lowest value	\(\lvert A \rvert = \dfrac{\max - \min}{2}\)
midline	the horizontal line halfway between the highest and the lowest value	\(y = D = \dfrac{\max + \min}{2}\)
period	the length of one full cycle	\(B = \dfrac{2\pi}{\text{period}}\)
phase shift	the horizontal shift of a sine or cosine curve	\(C\) in \(A\sin(B(t - C)) + D\) or \(A\cos(B(t - C)) + D\)
mirror image at the midline	the graph flipped upside down about its midline \(y = D\)	\(2D - f(t)\)

In set-builder notation, \(\{x \in \mathbb{R} \mid \dots\}\) and \(\{x \in \mathbb{R} : \dots\}\) mean the same.

Set B: Practice Mock (90 minutes, 50 points)

Problem 1: Meal Kits from a Start-Up [20 pts. total]

A start-up delivers meal kits (boxes with recipes and the measured ingredients for a few meals) to households in a city. The number of meal kits ordered per week, \(t\) weeks after the launch, is modelled by the logistic function

\[N(t) = \frac{L}{1 + A \cdot e^{-kt}}\]

in QU, with \(A, k > 0\). In the week of the launch (\(t = 0\)), the households order 3 QU, eight weeks later 21 QU. The start-up expects that the weekly orders will never exceed the saturation level of 24 QU. Round \(k\) to four decimal places, and orders and times to two decimal places.

(1 QU = 1,000 meal kits)

1. Explain why \(L = 24\), and determine \(A\) and \(k\) from the orders in the weeks \(t = 0\) and \(t = 8\). Give \(k\) as an exact term and as a rounded value.

(for verification: \(N(t) = \frac{24}{1 + 7 \cdot e^{-kt}}\) with \(k = \frac{\ln 7}{4} \approx 0.4865\))

[5 pts.]

2. Show that four weeks after the launch, the weekly orders are exactly half of the saturation level, and that they grow by the same amount from \(t = 0\) to \(t = 4\) as from \(t = 4\) to \(t = 8\). Compute the weekly orders twelve weeks after the launch in meal kits.

[4 pts.]

3. The start-up’s kitchen can pack at most 23 QU per week. Compute exactly and rounded the point in time at which the weekly orders reach 23 QU. Determine the first week after the launch in which the kitchen cannot pack all ordered meal kits.

[4 pts.]

4. Determine what happens to the weekly orders in the long run (\(t \to \infty\)) and give the horizontal asymptote of \(N\). Explain why the orders stay below 24 QU at every point in time, and interpret this for the start-up.

[3 pts.]

5. Sketch the graph of \(N\) for \(0 \leq t \leq 16\) together with the saturation level, and mark the three phases of logistic growth in your sketch. Describe each phase for the meal kits in one sentence.

[4 pts.]

CautionSolutions

a) Saturation level. Since \(k > 0\), the term \(e^{-kt}\) becomes arbitrarily small for large \(t\). The denominator \(1 + A \cdot e^{-kt}\) approaches 1 from above, so \(N(t)\) approaches \(L\) from below. The saturation level is therefore \(L = 24\).

Initial value. With \(e^0 = 1\):

\[N(0) = \frac{24}{1 + A} = 3 \iff 1 + A = 8 \iff A = 7\]

Growth constant. Use the second data point and multiply by the denominator (it is positive):

\[\frac{24}{1 + 7e^{-8k}} = 21 \iff 1 + 7e^{-8k} = \frac{24}{21} = \frac{8}{7} \iff 7e^{-8k} = \frac{1}{7} \iff e^{-8k} = \frac{1}{49}\]

Take the natural logarithm of both sides, with \(\ln(e^u) = u\) and \(\ln\left(\frac{1}{49}\right) = -\ln(49) = -2\ln 7\):

\[-8k = -2\ln 7 \iff k = \frac{\ln 7}{4} \approx 0.4865\]

\[N(t) = \frac{24}{1 + 7e^{-0.4865t}}\]

Check: \(N(8) = \frac{24}{1 + 7 \cdot \frac{1}{49}} = \frac{24}{8/7} = 21\) ✓

Grading: 1 Pt. \(L = 24\) with reason, 1 Pt. \(A = 7\), 1 Pt. equation \(e^{-8k} = \frac{1}{49}\), 1 Pt. logarithm, 1 Pt. \(k\) exactly and rounded.

b) From a), \(4k = \ln 7\), so \(e^{-4k} = e^{-\ln 7} = \frac{1}{7}\). (Or: \(e^{-4k} = \sqrt{e^{-8k}} = \sqrt{\frac{1}{49}} = \frac{1}{7}\).)

\[N(4) = \frac{24}{1 + 7 \cdot \frac{1}{7}} = \frac{24}{2} = 12 \quad \checkmark\]

Four weeks after the launch, the weekly orders are exactly half of the saturation level.

Increases: \(N(4) - N(0) = 12 - 3 = 9\) QU and \(N(8) - N(4) = 21 - 12 = 9\) QU ✓. The weekly orders grow by 9,000 meal kits in both periods.

Twelve weeks. With \(e^{-12k} = \left(e^{-4k}\right)^3 = \frac{1}{343}\):

\[N(12) = \frac{24}{1 + \frac{7}{343}} = \frac{24}{1 + \frac{1}{49}} = \frac{24 \cdot 49}{50} = 23.52 \text{ QU} = 23{,}520 \text{ meal kits per week}\]

Grading: 1 Pt. \(e^{-4k} = \frac{1}{7}\), 1 Pt. \(N(4) = 12\) as half of 24, 1 Pt. both increases 9 QU, 1 Pt. \(N(12)\) in meal kits.

c) Orders of 23 QU. Multiply by the denominator:

\[\frac{24}{1 + 7e^{-kt}} = 23 \iff 24 = 23 + 161e^{-kt} \iff 161e^{-kt} = 1 \iff e^{-kt} = \frac{1}{161}\]

Take the logarithm:

\[-kt = \ln\left(\frac{1}{161}\right) = -\ln 161 \iff t = \frac{\ln 161}{k} = \frac{4\ln 161}{\ln 7} \approx 10.45\]

First whole week: the denominator \(1 + 7e^{-kt}\) becomes smaller as \(t\) grows, so \(N\) grows, and the orders are above 23 QU for every \(t > 10.45\). \(t\) counts whole weeks (\(t = 0\) is the week of the launch), so the first such week is \(t = 11\), i.e. week 11 after the launch. From then on, the kitchen cannot pack all ordered meal kits.

Check:

• \(t = 10\): \(N(10) = \frac{24}{1 + 7e^{-10k}} \approx 22.77 < 23\)
• \(t = 11\): \(N(11) = \frac{24}{1 + 7e^{-11k}} \approx 23.23 > 23\) ✓

Grading: 1 Pt. equation \(e^{-kt} = \frac{1}{161}\), 1 Pt. logarithm and exact result, 1 Pt. rounded value, 1 Pt. week 11 with the check at \(t = 10\) and \(t = 11\).

d) For \(t \to \infty\): \(-kt \to -\infty\), so \(e^{-kt} \to 0\) and therefore

\[N(t) \to \frac{24}{1 + 7 \cdot 0} = 24\]

Horizontal asymptote: \(y = 24\).

Always below: since \(7e^{-kt} > 0\) for every \(t\), the denominator is always greater than 1, so \(N(t) < 24\).

Interpretation: in the long run, the households order almost 24 QU \(= 24{,}000\) meal kits per week, but the saturation level is never reached.

Grading: 1 Pt. \(N(t) \to 24\) with reason \(e^{-kt} \to 0\), 1 Pt. asymptote \(y = 24\) and interpretation, 1 Pt. \(N(t) < 24\) for every \(t\) with reason.

e) Table of values (\(N(t) = \frac{24}{1 + 7e^{-kt}}\), rounded to two decimal places):

\(t\)	0	2	4	6	8	10	12	16
\(N(t)\)	3	6.58	12	17.42	21	22.77	23.52	23.93

Figure 1: Graph of the weekly orders N(t) with its saturation level and the three phases

Three phases:

1. Slow start (about the first 2 weeks): only a few households know the service. From \(t = 0\) to \(t = 2\), the weekly orders rise by only about 3.58 QU.
2. Rapid growth (about weeks 2 to 6): word of mouth brings many new customers, and around \(t = 4\) the orders pass half of the saturation level (b)). From \(t = 2\) to \(t = 6\), they rise by about 10.8 QU.
3. Saturation (from about week 6 on): most interested households already order, and the orders approach 24 QU. From \(t = 12\) to \(t = 16\), they rise by only about 0.41 QU.

Note: the phases have no sharp boundaries. Other sensible approximate times are also correct.

Grading: 1 Pt. labeled axes and correct points, 1 Pt. S-shaped curve that stays below the asymptote \(y = 24\), 1 Pt. three phases marked with approximate times, 1 Pt. one sentence per phase.

Problem 2: A Cubic and a Rational Function [18 pts. total]

The figure shows the graph of \(f(x) = -x^3 + 2x^2 + 7x + 4\). Parts d) to f) also use the rational function \(g(x) = \dfrac{f(x)}{x^2 - 5x + 4}\).

Figure 2: Graph of the polynomial function f

1. Explain with the degree and the leading coefficient of \(f\) how the graph behaves for \(x \to -\infty\) and for \(x \to \infty\). Compute the \(y\)-intercept.

[2 pts.]

2. Determine all zeros of \(f\) with the Rational Root Theorem and polynomial long division, and give their multiplicities.

(for verification: \(f(x) = -(x + 1)^2(x - 4)\))

[4 pts.]

3. Determine the sets \(\{x \in \mathbb{R} \mid f(x) < 0\}\) and \(\{x \in \mathbb{R} \mid f(x) \leq 0\}\) with the factored form of \(f\). Explain why \(x = -1\) belongs to only one of the two sets.

[3 pts.]

4. Determine the domain of \(g\) and the hole. Investigate the asymptotic behavior of \(g\), and determine the \(x\) and \(y\) intercepts.

(for verification: \(g(x) = -\dfrac{(x + 1)^2}{x - 1}\) for all \(x\) in the domain, oblique asymptote \(y = -x - 3\))

[5 pts.]

5. Investigate on which side of the oblique asymptote the graph of \(g\) lies for \(x \to \infty\) and for \(x \to -\infty\).

[2 pts.]

6. Graph \(g\) directly into the figure with your results from d) and e). Include both asymptotes and mark the hole with an open circle.

[2 pts.]

CautionSolutions

a) \(f\) has degree 3 (odd) and the leading coefficient \(-1 < 0\). Therefore:

\[x \to -\infty: \; f(x) \to +\infty, \qquad x \to \infty: \; f(x) \to -\infty\]

The graph comes from the top left and ends at the bottom right.

\(y\)-intercept: \(f(0) = 4\), i.e. \((0, 4)\).

Grading: 1 Pt. both ends with the degree and the sign of the leading coefficient, 1 Pt. \((0, 4)\).

b) Rational Root Theorem: the leading coefficient is \(-1\), so the candidates are the divisors of 4: \(\pm 1, \pm 2, \pm 4\). The figure suggests \(x = -1\) and \(x = 4\).

Test:

\[f(-1) = 1 + 2 - 7 + 4 = 0 \quad \checkmark \qquad f(4) = -64 + 32 + 28 + 4 = 0 \quad \checkmark\]

So \((x - 4)\) is a factor of \(f(x)\).

Polynomial long division by \((x - 4)\) (divide, multiply, subtract):

\[\begin{array}{l} (-x^3 + 2x^2 + 7x + 4) : (x - 4) = -x^2 - 2x - 1 \\ \underline{-(-x^3 + 4x^2)} \\ \phantom{(-x^3}\;\; -2x^2 + 7x \\ \phantom{(-x^3}\;\; \underline{-(-2x^2 + 8x)} \\ \phantom{(-x^3 + 2x^2}\;\; -x + 4 \\ \phantom{(-x^3 + 2x^2}\;\; \underline{-(-x + 4)} \\ \phantom{(-x^3 + 2x^2 - x + }\;\; 0 \end{array}\]

Factor the quadratic as a perfect square: \(-x^2 - 2x - 1 = -(x^2 + 2x + 1) = -(x + 1)^2\).

\[f(x) = -(x + 1)^2(x - 4)\]

Check: \(-(x^2 + 2x + 1)(x - 4) = -(x^3 - 4x^2 + 2x^2 - 8x + x - 4) = -x^3 + 2x^2 + 7x + 4\) ✓

Zeros (zero-product rule): \(x_1 = -1\) (multiplicity 2) and \(x_2 = 4\) (multiplicity 1).

Grading: 1 Pt. candidates and a zero found by testing, 2 Pts. long division, 1 Pt. factored form with the zeros and their multiplicities.

c) Sign. In \(f(x) = -(x + 1)^2(x - 4)\), the factor \((x + 1)^2\) is zero at \(x = -1\) and positive for every other \(x\). So for \(x \neq -1\), \(f(x)\) has the sign of \(-(x - 4)\), and \(f(x) < 0 \iff x - 4 > 0\):

\[\{x \in \mathbb{R} \mid f(x) < 0\} = (4; \infty)\]

For \(f(x) \leq 0\), the two zeros \(x = -1\) and \(x = 4\) are added:

\[\{x \in \mathbb{R} \mid f(x) \leq 0\} = \{-1\} \cup [4; \infty)\]

\(x = -1\) is a double zero: \(f(-1) = 0\), but \(f\) is positive on both sides of it. The graph only touches the \(x\)-axis there from above, so \(x = -1\) is a single point of the second set and does not belong to the first.

Grading: 1 Pt. sign of \(f\) from the factored form, 1 Pt. \((4; \infty)\), 1 Pt. \(\{-1\} \cup [4; \infty)\) with the reason for \(x = -1\).

d) Domain. The denominator must not be zero:

\[x^2 - 5x + 4 = (x - 1)(x - 4) = 0 \iff x = 1 \text{ or } x = 4, \qquad D_g = \mathbb{R} \setminus \{1, 4\}\]

Hole. Simplify with b):

\[g(x) = \frac{-(x + 1)^2(x - 4)}{(x - 1)(x - 4)} = -\frac{(x + 1)^2}{x - 1} \quad \text{for } x \neq 1, \, 4 \quad \checkmark\]

At \(x = 4\) the factor \((x - 4)\) cancels. The simplified term gives \(-\frac{25}{3}\), so the hole is at \(\left(4, -\frac{25}{3}\right) \approx (4, -8.33)\).

Vertical asymptote: \(x = 1\) (the factor \((x - 1)\) stays in the denominator, and the numerator is \(-4 \neq 0\) there). Near \(x = 1\), the numerator is close to \(-4\) and the denominator close to 0:

\[g(x) \to +\infty \text{ for } x \to 1 \text{ from the left}, \qquad g(x) \to -\infty \text{ for } x \to 1 \text{ from the right}\]

Behavior for \(x \to \pm\infty\): the numerator has degree 2, the denominator degree 1. So there is no horizontal asymptote, but an oblique one. Divide the numerator \((x + 1)^2 = x^2 + 2x + 1\) by \(x - 1\):

\[\begin{array}{l} (x^2 + 2x + 1) : (x - 1) = x + 3 \;\text{ remainder } 4 \\ \underline{-(x^2 - x)} \\ \phantom{(x^2}\;\; 3x + 1 \\ \phantom{(x^2}\;\; \underline{-(3x - 3)} \\ \phantom{(x^2 + 3x}\;\; 4 \end{array}\]

So \(\frac{(x + 1)^2}{x - 1} = x + 3 + \frac{4}{x - 1}\) and

\[g(x) = -x - 3 - \frac{4}{x - 1}\]

For \(x \to \pm\infty\), the remainder term \(\frac{4}{x - 1} \to 0\): oblique asymptote \(y = -x - 3\).

Intercepts. \(-(x + 1)^2 = 0\) gives \(x = -1\), which lies in the domain, so the only \(x\)-intercept is \((-1, 0)\). \(x = 4\) is not a zero of \(g\), although \(f(4) = 0\): it is not in the domain, and the graph has a hole there. \(y\)-intercept: \(g(0) = -\frac{1}{-1} = 1\), i.e. \((0, 1)\).

Grading: 1 Pt. domain, 1 Pt. cancelling and hole, 1 Pt. vertical asymptote \(x = 1\) with the behavior on both sides, 1 Pt. long division and oblique asymptote \(y = -x - 3\), 1 Pt. \(x\)-intercept (with \(x = 4\) excluded) and \(y\)-intercept.

e) The difference between \(g\) and the asymptote is the remainder term: \(g(x) - (-x - 3) = -\frac{4}{x - 1}\).

• \(x \to \infty\): \(x - 1 > 0\), so the difference is negative and the graph lies below the asymptote (e.g. \(g(5) = -9 < -8\)).
• \(x \to -\infty\): \(x - 1 < 0\), so the difference is positive and the graph lies above the asymptote (e.g. \(g(-4) = 1.8 > 1\)).

Grading: 1 Pt. below for \(x \to \infty\) with reason, 1 Pt. above for \(x \to -\infty\) with reason.

f) Table of values (from \(g(x) = -\frac{(x + 1)^2}{x - 1}\)):

\(x\)	\(-4\)	\(-3\)	\(-2\)	\(-1\)	\(0\)	\(0.5\)	\(1.5\)	\(2\)	\(3\)	\(5\)	\(7\)
\(g(x)\)	\(1.8\)	\(1\)	\(0.33\)	\(0\)	\(1\)	\(4.5\)	\(-12.5\)	\(-9\)	\(-8\)	\(-9\)	\(-10.67\)

Figure 3: Graph of f with the rational function g and its asymptotes

Grading: 1 Pt. both asymptotes and both branches with the correct behavior at them, 1 Pt. zero \((-1, 0)\) where the graph touches the \(x\)-axis, \(y\)-intercept and open circle at the hole.

Problem 3: Tyre Changes at a Car Workshop [12 pts. total]

A car workshop changes summer tyres for winter tyres and back. The number of tyre changes per week follows a pattern with two seasons per year: it is highest on 1 April and on 1 October with 285 tyre changes per week, and lowest on 1 July and on 1 January with 45 tyre changes per week. The workshop models the number of tyre changes per week by a sinusoidal function \(T(t)\), where \(t\) is the time in months after 1 March (\(t = 0\) stands for 1 March, \(t = 1\) for 1 April).

1. Give the amplitude and the midline of the model. Determine the period and the parameter \(B\).

[2 pts.]

2. Give \(T(t)\) in the form \(A\cos(B(t - C)) + D\) and give a reason for your choice of \(C\). Give \(T(t)\) also in the form \(A\sin(B(t - C)) + D\) with \(0 \leq C < 6\) and check this form at \(t = 1\).

(for verification: \(T(t) = 165 + 120\cos\left(\frac{\pi}{3}(t - 1)\right)\))

[4 pts.]

3. The workshop’s own mechanics can do at most as many tyre changes per week as the model gives for the middle of March (\(t = 0.5\)). In all weeks with more tyre changes, the workshop hires a temporary mechanic. Determine exactly the time intervals of the year in which the workshop needs a temporary mechanic, and give them as dates.

Hint: in \([-\pi; \pi]\), \(\cos(u) = \frac{\sqrt{3}}{2}\) exactly for \(u = -\frac{\pi}{6}\) and \(u = \frac{\pi}{6}\).

[4 pts.]

4. Argue whether the following statements are true or false.

1. If the lowest number stays at 45 tyre changes per week and the highest number rises to 325, only the amplitude of the model changes.
2. \(T(t + 3) = 330 - T(t)\) for every \(t\).

[2 pts.]

CautionSolutions

a)

\[\text{Amplitude: } |A| = \frac{\max - \min}{2} = \frac{285 - 45}{2} = 120\]

\[\text{Midline: } D = \frac{\max + \min}{2} = \frac{285 + 45}{2} = 165, \text{ i.e. } y = 165\]

Period: from the maximum on 1 April (\(t = 1\)) to the next maximum on 1 October (\(t = 7\)) are 6 months, so

\[B = \frac{2\pi}{6} = \frac{\pi}{3}\]

Check: the minimum on 1 July (\(t = 4\)) lies half a period after the maximum at \(t = 1\) ✓

Grading: 1 Pt. amplitude and midline, 1 Pt. period and \(B\).

b) Cosine form. A cosine starts at its maximum. The first maximum is at \(t = 1\) (1 April), so the cosine is shifted 1 month to the right: \(C = 1\).

\[T(t) = 165 + 120\cos\left(\frac{\pi}{3}(t - 1)\right)\]

Check: \(T(1) = 165 + 120 = 285\) ✓ and \(T(4) = 165 + 120\cos(\pi) = 45\) ✓

Sine form. A sine starts on the midline and rises. \(T\) crosses its midline upwards a quarter period (1.5 months) after the minimum at \(t = 4\), i.e. at \(t = 5.5\) (in the middle of August). So \(C = 5.5\):

\[T(t) = 165 + 120\sin\left(\frac{\pi}{3}(t - 5.5)\right)\]

(The upward crossing a quarter period before the maximum, at \(t = 1 - 1.5 = -0.5\), lies outside \(0 \leq C < 6\). One period later it is at \(-0.5 + 6 = 5.5\).)

Check at \(t = 1\): \(165 + 120\sin\left(\frac{\pi}{3} \cdot (-4.5)\right) = 165 + 120\sin\left(-\frac{3\pi}{2}\right) = 165 + 120 \cdot 1 = 285\) ✓. A full turn gives \(-\frac{3\pi}{2} + 2\pi = \frac{\pi}{2}\), the point \((0, 1)\) at the top of the unit circle.

Grading: 1 Pt. cosine form, 1 Pt. reason for \(C = 1\), 1 Pt. upward midline crossing at \(t = 5.5\) with \(0 \leq C < 6\), 1 Pt. sine form with check at \(t = 1\).

c) Capacity of the own mechanics (with the hint, \(\cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)):

\[T(0.5) = 165 + 120\cos\left(-\frac{\pi}{6}\right) = 165 + 120 \cdot \frac{\sqrt{3}}{2} = 165 + 60\sqrt{3} \approx 268.92\]

Condition:

\[165 + 120\cos\left(\frac{\pi}{3}(t - 1)\right) > 165 + 60\sqrt{3} \iff \cos\left(\frac{\pi}{3}(t - 1)\right) > \frac{\sqrt{3}}{2}\]

Dividing by the positive number 120 keeps the inequality sign.

First season. With \(u = \frac{\pi}{3}(t - 1)\), the hint gives the boundaries \(u = -\frac{\pi}{6}\) and \(u = \frac{\pi}{6}\). Between them, the cosine passes its maximum 1 at \(u = 0\), so it is greater than \(\frac{\sqrt{3}}{2}\) there:

\[-\frac{\pi}{6} < \frac{\pi}{3}(t - 1) < \frac{\pi}{6} \iff -\frac{1}{2} < t - 1 < \frac{1}{2} \iff 0.5 < t < 1.5\]

Second season. The pattern repeats after one period: \(u\) grows by a full turn \(2\pi\), and \(t\) by 6 months:

\[0.5 + 6 < t < 1.5 + 6 \iff 6.5 < t < 7.5\]

\[t \in (0.5; 1.5) \cup (6.5; 7.5)\]

Check: \(T(1) = T(7) = 285 > 268.92\) ✓ and \(T(4) = 45 < 268.92\) ✓

Dates: the workshop needs a temporary mechanic from the middle of March to the middle of April and from the middle of September to the middle of October, i.e. for 2 months per year, around the two peaks of the tyre seasons.

Grading: 1 Pt. \(T(0.5) = 165 + 60\sqrt{3}\), 1 Pt. inequality for the cosine, 1 Pt. first interval from the hint, 1 Pt. second interval with the period and dates.

d)

1. False. New amplitude \(\frac{325 - 45}{2} = 140\) and new midline \(\frac{325 + 45}{2} = 185\). The midline moves up by 20 as well. With only a larger amplitude, the lowest value would fall to \(165 - 140 = 25\).
2. True. Three months are half a period: \(T(t + 3) = 165 + 120\cos\left(\frac{\pi}{3}(t - 1) + \pi\right)\). Half a turn on the unit circle leads to the opposite point, so \(\cos(u + \pi) = -\cos(u)\) and \(T(t + 3) = 165 - 120\cos\left(\frac{\pi}{3}(t - 1)\right) = 330 - T(t)\). Three months later, the value is the mirror image at the midline \(y = 165\), e.g. \(T(1) = 285\) and \(T(4) = 45 = 330 - 285\).

Grading: 1 Pt. per statement (decision and reason).

Exam Preparation Checklist

Before the exam, ensure you can:

• Describe the end behavior of a polynomial with its degree and leading coefficient, and read zeros and multiplicities from the factored form or from a graph
• Find a zero with the Rational Root Theorem or check a given one, divide by polynomial long division and factor completely
• Use the Intermediate Value Theorem to show that a zero lies in an interval
• Determine the domain, the zeros, the holes and all asymptotes of a rational function, and decide on which side of an oblique asymptote the graph lies
• Sketch a model with its horizontal asymptote, and graph a rational function with its asymptotes and hole directly into a given figure
• Explain what the exponent of a power function says, e.g. that costs grow less than proportionally
• Set up exponential models from data, switch between \(b^t\) and \(e^{kt}\), and compute doubling times and half-lives
• Solve exponential equations and inequalities with logarithms, and flip the sign when you divide by a negative logarithm
• Show a property for every \(t\) by splitting a power such as \(e^{k(t + T)} = e^{kt} \cdot e^{kT}\)
• Determine the parameters of a model such as \(L - a \cdot e^{-kt}\) or \(\frac{L}{1 + A e^{-kt}}\) when the task gives the form, and interpret its behavior for \(t \to \infty\)
• Convert between degrees and radians and use the exact values of sine and cosine
• Determine the amplitude, the midline, the period, \(B\) and the phase shift of a sinusoidal model, and switch between the sine and the cosine form
• Solve inequalities with sinusoids from the basic solutions given in a hint, also over more than one period

Remember: Show all work clearly, label units, and always interpret your results in the business context!