Interpreting and Comparing Functions: Algebra 2 Math Worksheet Answer Key – PDF Download

Algebra 2 Function Worksheet Answer Key: problems 1–5

Problem 1: Key features from a graph

1. The maximum is 75°F at \(h=12\) (noon), since \(60+15=75\).
2. The minimum is 45°F at \(h=0\) and \(h=24\) (midnight), since \(60-15=45\).
3. One full cycle takes 24 hours. The midline is 60°F, halfway between 45 and 75.
4. Increasing on \(0 < h < 12\) and decreasing on \(12 < h < 24\).
5. \(T(6)=60\) and \(T(12)=75\), so \(\frac{75-60}{12-6}=2.5\): 2.5°F per hour.

Problem 2: Domain from context

1. The length is \(40-w\), and both the width and the length must be positive, so \(w\) is greater than 0 and less than 40.
2. The domain is \(0 < w < 40\) feet.
3. The zeros are \(w=0\) and \(w=40\), so the axis of symmetry is halfway at \(w=20\). Then \(A(20)=20(20)=400\). The greatest area is 400 square feet with a width of 20 ft (a square).

Problem 3: Discrete domain

1. Tickets are whole numbers from 0 to 150, so the domain is \(n=0, 1, 2\) and so on, up to 150.
2. \(C(40)=8.5(40)=340\) and \(C(150)=8.5(150)=1275\). The revenue is $340 and $1,275.
3. The graph is a set of separate points on the line \(y=8.5x\), because you cannot sell part of a ticket.

Problem 4: Rational model in context

1. \(C(8)=62.50\), \(C(20)=25\), \(C(50)=10\) dollars.
2. It is decreasing: more members means a smaller share for each.
3. The horizontal asymptote is \(C=0\). The share gets closer to $0 as the group grows but never reaches it.
4. The number of members is a positive whole number, so the domain is \(x=1,2,3\) and so on; the vertical asymptote at \(x=0\) is not part of the situation.

Problem 5: Exponential key features

1. \(V(0)=1200\): the laptop cost $1,200 new.
2. Decreasing, because the base 0.8 is between 0 and 1 (a 20% loss each year).
3. The asymptote is \(V=0\): the value approaches $0 but never becomes negative.
4. \(V(3)=1200(0.512)=614.4\), so $614.40.
5. \(V(3)=614.40\) is still above 500, and \(V(4)=1200(0.4096)=491.52\) is below. The answer is year 4.

Algebra 2 Function Worksheet Answer Key: problems 6–10

Problem 6: Logarithmic model

1. \(-\log 10^{-7}=7\), \(-\log 10^{-3}=3\) and \(-\log 10^{-4}=4\). The pH values are 7, 3 and 4.
2. A concentration must be positive, so the domain is \(c>0\) (the vertical asymptote is \(c=0\)).
3. The function is decreasing. Multiplying \(c\) by 10 adds 1 to \(\log c\), so the pH drops by 1.

Problem 7: Average rate of change from equations

1. \(f(4)=4\) and \(f(1)=-2\), so \(\frac{4-(-2)}{4-1}=\frac{6}{3}=2\). The rate is 2.
2. \(g(3)=27\) and \(g(-1)=-1\), so \(\frac{27-(-1)}{3-(-1)}=\frac{28}{4}=7\). The rate is 7.
3. \(h(4)=16\) and \(h(1)=2\), so \(\frac{16-2}{4-1}=\frac{14}{3}\). The rate is \(\frac{14}{3}\approx 4.67\).

Problem 8: Average rate of change from a table

1. \(\frac{1.2-0.3}{6-2}=\frac{0.9}{4}=0.225\) miles per minute.
2. \(\frac{0.7-0}{4-0}=0.175\) mi/min on \([0,4]\) and \(\frac{1.9-0.7}{8-4}=0.3\) mi/min on \([4,8]\).
3. The averages rise from 0.175 to 0.3 miles per minute, so Sofia speeds up as the run goes on.

Problem 9: Rate of change and features from a graph

1. \(\frac{4-0}{1-0}=4\), \(\frac{0-4}{3-1}=-2\) and \(\frac{4-0}{4-0}=1\).
2. Since \(g(x)=x(x-3)^{2}\), the zeros are 0 and 3 (the graph touches the axis at 3). It increases for \(x < 1\) and \(x > 3\), and decreases for \(1 < x < 3\).
3. The relative maximum is 4 at \(x=1\) and the relative minimum is 0 at \(x=3\).

Problem 10: Compare average rates

1. For \(f\): \(\frac{16-1}{4}=3.75\). For \(g\): \(\frac{16-0}{4}=4\). On \([0,4]\), \(g\) changes faster.
2. For \(f\): \(\frac{64-16}{2}=24\). For \(g\): \(\frac{24-16}{2}=4\). On \([4,6]\), \(f\) changes much faster.
3. The linear rate stays at 4, but the exponential rate keeps growing and eventually passes it.

Algebra 2 Function Worksheet Answer Key: problems 11–15

Problem 11: Zeros from factored form

1. \(f(x)=(x-2)(x-5)\), so the zeros are \(x=2\) and \(x=5\).
2. The axis is halfway between the zeros: \(x=\frac{2+5}{2}=3.5\).
3. \(f(3.5)=(1.5)(-1.5)=-2.25\). The minimum value is \(-2.25\).

Problem 12: Maximum from factored form

1. \(g(x)=-2(x^{2}-4x-5)=-2(x-5)(x+1)\), so the zeros are \(x=5\) and \(x=-1\).
2. The axis is \(x=\frac{5+(-1)}{2}=2\), and \(g(2)=-8+16+10=18\). The maximum value is 18.
3. \(g(0)=10\), so the y-intercept is 10.

Problem 13: Complete the square

1. \(x^{2}+6x+9-9+5\) gives \(f(x)=(x+3)^{2}-4\). The vertex is \((-3,-4)\). Setting \((x+3)^{2}=4\) gives \(x+3=\pm 2\), so the zeros are \(-5\) and \(-1\).
2. \(x^{2}-10x+25-25+19\) gives \(h(x)=(x-5)^{2}-6\). The vertex is \((5,-6)\). Setting \((x-5)^{2}=6\) gives zeros \(x=5\pm\sqrt{6}\).

Problem 14: Complete the square with a leading coefficient

1. \(2(x^{2}-6x)+7=2(x-3)^{2}-18+7\), so \(f(x)=2(x-3)^{2}-11\). The parabola opens up, so \((3,-11)\) is a minimum. From \((x-3)^{2}=\frac{11}{2}\), the zeros are \(x=3\pm\frac{\sqrt{22}}{2}\).
2. \(-3(x^{2}-2x)+9=-3(x-1)^{2}+3+9\), so \(g(x)=-3(x-1)^{2}+12\). It opens down, so \((1,12)\) is a maximum. From \((x-1)^{2}=4\), the zeros are \(x=3\) and \(x=-1\).

Problem 15: Profit from a quadratic model

1. \(P(x)=-2(x^{2}-30x+125)=-2(x-5)(x-25)\), so the break-even prices are $5 and $25.
2. \(-2(x^{2}-30x+225)+450-250\) gives \(P(x)=-2(x-15)^{2}+200\).
3. The maximum occurs at \(x=15\): a price of $15 gives a profit of $200.
4. \(P(0)=-250\): giving tickets away would lose $250 (the fixed costs).

Algebra 2 Function Worksheet Answer Key: problems 16–20

Problem 16: Compare two quadratics

1. \(f(x)=x^{2}-6x+9-9+5=(x-3)^{2}-4\), so the vertex is \((3,-4)\).
2. The minimum of \(f\) is \(-4\) and the minimum of \(g\) is 2 (vertex \((1,2)\)). The function \(f\) has the lesser minimum.
3. The function \(f\) has zeros \(x=1\) and \(x=5\) because \((x-3)^{2}=4\). The graph of \(g\) opens up with minimum 2, so it never reaches 0: only \(f\) has real zeros.
4. \(f(0)=5\) and \(g(0)=3\) (from the graph), so \(f\) has the greater y-intercept.

Problem 17: Compare a table and an equation

1. Each value is twice the one before, so \(A(x)=3(2)^{x}\) and \(A(4)=48\).
2. \(B(4)=5(4)+4=24\), so \(A\) is greater (48 versus 24).
3. For \(A\): \(\frac{24-3}{3}=7\). For \(B\): \(\frac{B(3)-B(0)}{3}=\frac{19-4}{3}=5\). The rate of \(A\) is greater.

Problem 18: Compare two projectiles

1. \(h_{A}(t)=-16(t^{2}-2t)+6=-16(t-1)^{2}+22\). The maximum is 22 feet at \(t=1\) second.
2. The table is symmetric about \(t=1.5\) with a top value of 40, so Ball B reaches 40 feet. Ball B goes higher.
3. \(h_{A}(0)=6\) and Ball B starts at 4, so Ball A starts higher.
4. Ball A: \(\frac{22-6}{1}=16\) ft/s. Ball B: \(\frac{36-4}{1}=32\) ft/s. Ball B is rising faster at first.

Problem 19: Average cost model

1. \(A(10)=\frac{250}{10}=25\), \(A(40)=\frac{400}{40}=10\) and \(A(100)=\frac{700}{100}=7\) dollars.
2. \(\frac{7-25}{100-10}=-0.2\): on average the cost per sticker falls $0.20 for each extra sticker between 10 and 100.
3. \(A(n)=5+\frac{200}{n}\), so the horizontal asymptote is \(A=5\). The $200 setup cost is spread over more stickers, so the average approaches the $5 per-sticker cost.
4. \(5+\frac{200}{n}=6\) gives \(n=200\), so 200 stickers.

Problem 20: Ferris wheel heights

1. The lowest height is 5 ft at \(t=0\) and \(t=60\); the highest is 45 ft at \(t=30\).
2. The period is 60 minutes: that is the time for one full turn.
3. \(h(10)=25-20\cos\frac{\pi}{3}=15\) gives \(\frac{15-5}{10}=1\) ft/min on \([0,10]\). \(h(20)=25-20\cos\frac{2\pi}{3}=35\) gives \(\frac{35-15}{10}=2\) ft/min on \([10,20]\). The seat rises faster on \([10,20]\).

Algebra 2 Function Worksheet Answer Key: problems 21–22

Problem 21: Open box volume

1. The side \(x\) must be positive, and \(9-2x\) must be positive, so the domain is \(0 < x < 4.5\).
2. The greatest table value is 81 cubic inches at \(x=1.5\). (The true maximum is slightly above 81, near \(x=1.7\).)
3. \(\frac{81-44}{1}=37\) in³ per inch, positive: the volume increases. \(\frac{54-81}{1.5}=-18\) in³ per inch, negative: the volume decreases.

Problem 22: Compare savings plans

1. Plan A: \(150+25(6)=300\) and \(150+25(7)=325\) dollars. Plan B: \(100(1.2)^{6}\approx 298.60\) and \(100(1.2)^{7}\approx 358.32\) dollars.
2. After week 6, A has $300 and B has $298.60. After week 7, B has $358.32 against A’s $325. Plan B first leads in week 7.
3. Plan A: \(\frac{300-150}{6}=25\) dollars per week. Plan B: \(\frac{298.60-100}{6}\approx 33.10\) dollars per week. Over 6 weeks, B’s average rate is greater even though A has slightly more money.

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