Each solution shows the work, with the final result in bold. When your interval or point differs, check the order of subtraction in rate-of-change problems, and check that you used strict inequalities for open intervals.
Problem 1: Intercepts of a line
For the \(y\)-intercept, \(f(0)=2(0)-6=-6\).
For the \(x\)-intercept, \(2x-6=0\), so \(2x=6\) and \(x=3\).
\(y\)-intercept \((0,-6)\); \(x\)-intercept \((3,0)\)
Problem 2: Intercepts of a parabola
\(g(0)=-9\), so the \(y\)-intercept is \((0,-9)\).
Solve \(x^{2}-9=0\): \(x^{2}=9\), so \(x=3\) or \(x=-3\).
\(y\)-intercept \((0,-9)\); \(x\)-intercepts \((-3,0)\) and \((3,0)\)
Problem 3: Features from a graph
The graph meets the \(x\)-axis at 1 and 3 and the \(y\)-axis at 3. The lowest point is \((2,-1)\). The axis of symmetry is the vertical line through the vertex; it is halfway between the intercepts 1 and 3.
\(x\)-intercepts \((1,0)\), \((3,0)\); \(y\)-intercept \((0,3)\); vertex \((2,-1)\), a minimum; axis \(x=2\)
Problem 4: Intercepts from a table
The column \(x=0\) gives \(f(0)=12\). The column with output 0 is \(x=3\). The outputs drop by 4 each step.
\(y\)-intercept \((0,12)\); \(x\)-intercept \((3,0)\); decreasing
Problem 5: Increasing and decreasing
Reading left to right, the graph rises to \((-3,3)\), falls to \((1,-3)\), then rises again.
Increasing: \(-5 < x < -3\) and \(1 < x < 5\). Decreasing: \(-3 < x < 1\). Relative maximum 3 at \(x=-3\); relative minimum \(-3\) at \(x=1\).
Problem 6: Positive and negative intervals
The graph meets the \(x\)-axis at \(-4\), 0 and 4. Between \(-4\) and 0 it is above the axis; between 0 and 4 it is below the axis.
\(x\)-intercepts \(-4\), 0, 4. Positive: \(-4 < x < 0\). Negative: \(0 < x < 4\).
Problem 7: Axis of symmetry and vertex
The axis is \(x=-\frac{b}{2a}=-\frac{-6}{2(1)}=3\). Then \(f(3)=9-18+5=-4\), so the vertex is \((3,-4)\).
The \(y\)-intercept is \((0,5)\). It is 3 units left of the axis, so the mirror point is 3 units right: \((6,5)\). Check: \(f(6)=36-36+5=5\).
Axis \(x=3\); vertex \((3,-4)\); symmetric point \((6,5)\)
Problem 8: Features of an absolute value function
The vertex is where \(x-2=0\): \((2,-3)\), a minimum. The \(y\)-intercept is \(f(0)=2-3=-1\). For \(x\)-intercepts, \(|x-2|=3\), so \(x-2=3\) or \(x-2=-3\), giving \(x=5\) or \(x=-1\).
Vertex \((2,-3)\); \(y\)-intercept \((0,-1)\); \(x\)-intercepts \((-1,0)\), \((5,0)\); decreasing for \(x < 2\); increasing for \(x > 2\); symmetric about \(x=2\)
Problem 9: Maximum of a quadratic
The axis is \(x=-\frac{4}{2(-1)}=2\) and \(f(2)=-4+8+1=5\). The parabola opens down, so the vertex \((2,5)\) is a maximum. The \(y\)-intercept is \(f(0)=1\).
Vertex \((2,5)\); maximum value 5; \(y\)-intercept \((0,1)\); increasing for \(x < 2\); decreasing for \(x > 2\)
Problem 10: End behavior
- Large \(x\): the outputs fall without bound. Very negative \(x\): the outputs rise without bound. Falls on the right, rises on the left.
- Large or very negative \(x\): \(x^{2}\) is large, so \(g(x)\) rises without bound at both ends. Rises on both ends.
- Large \(x\): \(3^{x}\) grows without bound. Very negative \(x\): \(3^{x}\) is a tiny positive number, like \(\frac{1}{27}\) for \(x=-3\). Grows on the right, approaches 0 on the left.
Problem 11: Average rate of change: squares
\(f(1)=1\) and \(f(4)=16\). Then \(\frac{16-1}{4-1}=\frac{15}{3}=5\).
5
Problem 12: A rate of change of zero
\(f(-1)=1+2=3\) and \(f(3)=9-6=3\). So \(\frac{3-3}{3-(-1)}=\frac{0}{4}=0\).
The function goes down and then up, so the changes cancel; the starting and ending outputs are equal.
0
Problem 13: Average rate of change from a graph
Read the points: \(f(0)=0\), \(f(1)=3\), \(f(2)=4\), \(f(3)=3\), \(f(5)=-5\).
\(\frac{4-0}{2-0}=2\); \(\frac{-5-4}{5-2}=\frac{-9}{3}=-3\); \(\frac{3-3}{3-1}=0\).
2, \(-3\) and 0
Problem 14: Rate of change of an exponential
\(f(1)=2\), \(f(3)=8\), \(f(5)=32\).
\(\frac{8-2}{3-1}=\frac{6}{2}=3\) and \(\frac{32-8}{5-3}=\frac{24}{2}=12\).
The rate of change is four times as large on the later interval, so the function grows faster and faster.
3 and 12
Problem 15: Equation versus table
For \(f\): rate 3, \(y\)-intercept \(-2\). For \(g\): the outputs rise 4 per step, so the rate is 4, and \(g(0)=1\).
At \(x=10\): \(f(10)=28\), and \(g(10)=1+4(10)=41\).
\(g\) has the greater rate (4 vs 3), the greater \(y\)-intercept (1 vs \(-2\)), and the greater value at 10 (41 vs 28).
Problem 16: Equation versus graph
For \(f\): the vertex is \((0,9)\), so the maximum is 9. \(f(0)=9\). Solve \(-x^{2}+9=0\): \(x=\pm 3\).
For \(g\): the maximum is the vertex height, 4. The \(y\)-intercept is \(-5\). The \(x\)-intercepts are 1 and 5.
\(f\) has the greater maximum (9 vs 4) and the greater \(y\)-intercept (9 vs \(-5\)). \(f\): \(x=-3,3\); \(g\): \(x=1,5\).
Problem 17: Ball height features
\(h(0)=5\): the ball starts 5 feet above the ground.
The axis is \(t=-\frac{64}{2(-16)}=2\), and \(h(2)=-64+128+5=69\).
\(\frac{h(2)-h(0)}{2-0}=\frac{69-5}{2}=32\) feet per second.
Starts at 5 ft; highest at 2 s, 69 ft; average rate 32 ft/s
Problem 18: Draining a tank
\(V(0)=120\): the tank starts with 120 gallons. Solve \(120-8t=0\): \(t=15\), so the tank is empty after 15 minutes.
The rate of change is \(-8\): the tank loses 8 gallons every minute.
\(V\)-intercept 120 gal; \(t\)-intercept 15 min; rate \(-8\) gal/min; domain \(0\le t\le 15\)
Problem 19: Tide chart
The graph is above the axis from \(t=0\) to \(t=6\). It rises from 0 to 3 hours and from 9 to 12 hours. The peak is \((3,4)\) and the lowest point is \((9,-4)\).
\(\frac{-4-4}{9-3}=\frac{-8}{6}=-\frac{4}{3}\) feet per hour.
Above dock: \(0 < t < 6\). Increasing: \(0 < t < 3\) and \(9 < t < 12\). Maximum 4 ft at 3 h; minimum \(-4\) ft at 9 h. Average rate \(-\frac{4}{3}\) ft/h.
Problem 20: Road trip speeds
\(\frac{230-0}{4-0}=57.5\) mph. \(\frac{180-62}{3-1}=\frac{118}{2}=59\) mph. \(\frac{230-118}{4-2}=\frac{112}{2}=56\) mph.
57.5 mph, 59 mph and 56 mph; the lowest speed was from \(t=2\) to \(t=4\).
Problem 21: Gym memberships
Gym A has rate \(\$20\) per month (plus a \(\$50\) fee). Gym B rises \(\$45\) per month, so \(C_B(m)=45m\).
After 1 month: A \(=70\), B \(=45\). After 2 months: A \(=90\), B \(=90\). After 6 months: A \(=20(6)+50=170\), B \(=45(6)=270\).
B has the greater rate (45 vs 20). Gym A is cheaper after more than 2 months, the same at 2 months, and costlier for 1 month.
Problem 22: Exponential versus linear growth
From 0 to 4: \(F\): \(\frac{16-1}{4}=3.75\); \(G\): \(\frac{40-0}{4}=10\).
From 0 to 6: \(F\): \(\frac{64-1}{6}=10.5\); \(G\): \(\frac{60-0}{6}=10\).
On \(0\) to \(4\), \(G\) is greater (10 vs 3.75). On \(0\) to \(6\), \(F\) is greater (10.5 vs 10): the exponential overtakes the line.
See every chapter on the Algebra 1 math page.