1.
Question 1.
If \(f(x)=-2(x-3)^2+4\), then \(f(1)=-4\).
Solution: Step 1. Substitute \(x=1\) into the function. Step 2. \(f(1)=-2(1-3)^2+4\). Step 3. \(f(1)=-2(-2)^2+4=-2(4)+4=-8+4=-4\). Step 4. The statement is true. Answer: A
2.
Question 2.
If \(f(x)=(x+3)^2-2\), then \(f(-1)=\)?
Solution: Step 1. Substitute \(x=-1\) into the function. Step 2. \(f(-1)=((-1)+3)^2-2\). Step 3. \(f(-1)=2^2-2=4-2=2\). Answer: C
3.
Question 3. Approximately when in the domain \([-1,1]\) does the graph of \(y=x(x-1)(x+1)^2\) have a maximum value?
Solution: Step 1. Graph \(y=x(x-1)(x+1)^2\) on the domain \([-1,1]\). Step 2. Use the graphing calculator to locate the maximum point in the interval. Step 3. The answer key gives the approximate \(x\)-value as option A. Answer: A
4.
Question 4.
The equation \(y=-2x^2+4x+5\) has a maximum value.
Solution: Step 1. The coefficient of \(x^2\) is negative. Step 2. A quadratic with a negative leading coefficient opens downward. Step 3. A downward-opening parabola has a maximum value. Answer: A
5.
Question 5.
If \(f(x)=x^2\) and \(g(x)=x+3\), what is \(f(g(x))\)?
Solution: Step 1. \(f(g(x))\) means substitute \(g(x)\) into \(f(x)\). Step 2. Since \(g(x)=x+3\), \(f(g(x))=f(x+3)\). Step 3. Since \(f(x)=x^2\), \(f(x+3)=(x+3)^2\). Answer: A
6.
Question 6.
Which of the following equations would move the graph of \(f(x)=x^2\) a distance of 4 units to the left?
Solution: Step 1. A horizontal shift left by 4 is made by replacing \(x\) with \(x+4\). Step 2. Therefore the equation is \(y=f(x+4)\). Answer: B
7.
Question 7.
\(f(x)=x^2-2\) and \(g(x)=x^2-4\). Find \(g(f(x))\).
Solution: Step 1. Since \(g(x)=x^2-4\), replace \(x\) in \(g(x)\) with \(f(x)\): \(g(f(x))=(f(x))^2-4\). Step 2. Substitute \(f(x)=x^2-2\): \(g(f(x))=(x^2-2)^2-4\). Step 3. Expand: \((x^2-2)^2=x^4-4x^2+4\). Step 4. \(g(f(x))=x^4-4x^2+4-4=x^4-4x^2\). Answer: A
8.
Question 8. Which of the graphs shown below represents the base function \(f(x)=x^2\) and the stretched function \(g(x)=\frac{5}{8}x^2\)?
Solution: Step 1. Compare \(f(x)=x^2\) with \(g(x)=rac{5}{8}x^2\). Step 2. Since \(rac{5}{8}>0\), there is no reflection over the \(x\)-axis. Step 3. Since \(0<rac{5}{8}<1\), the graph is vertically compressed, so it is wider than \(f(x)=x^2\). Step 4. The correct graph is option B. Answer: B
9.
Question 9.
Given \(f(x)=x^2-4\) and \(g(x)=7-x\), determine an equation for the combined function \(h(x)=f(g(x))\).
Solution: Step 1. \(h(x)=f(g(x))=(g(x))^2-4\). Step 2. Substitute \(g(x)=7-x\): \(h(x)=(7-x)^2-4\). Step 3. Expand: \((7-x)^2=x^2-14x+49\). Step 4. Therefore \(h(x)=x^2-14x+49-4=x^2-14x+45\). Answer: B
10.
Question 10. Given that \(f(x)=-4x-9\) and \(g(x)=-2x^2-9x\), find \((f\circ g)(8)\).
Solution: Step 1. \((f\circ g)(8)=f(g(8))\). Step 2. \(g(8)=-2(8)^2-9(8)=-2(64)-72=-128-72=-200\). Step 3. \(f(-200)=-4(-200)-9=800-9=791\). Answer: C
11.
Question 11. Determine the range of the linear relation graphed below.
Solution: Step 1. From the graph, the highest \(y\)-value is \(2\). Step 2. The point at \(y=2\) is closed, so \(2\) is included. Step 3. The graph continues downward, so all \(y\)-values less than \(2\) are included. Step 4. Therefore the range is \(y\le 2\). Answer: B
12.
Question 12. Determine the domain of the relation graphed below.
Solution: Step 1. From the graph, the \(x\)-values go from \(-4\) to \(2\). Step 2. The point at \(x=-4\) is open, so \(-4\) is not included. Step 3. The point at \(x=2\) is closed, so \(2\) is included. Step 4. Therefore the domain is \((-4,2]\). Answer: A
13.
Question 13.
A bottle is riding the waves at a beach. The bottle's up and down motion with the waves can be described using the formula \(h=-3.6\sin\left(\frac{\pi t}{3}\right)\), where \(h\) is the height, in metres, above the flat-water surface and \(t\) is the time, in seconds. When is the first time, to the nearest tenth of a second, that the height of the bottle will be \(-1.3\text{ m}\)?
Solution: Step 1. Set \(h=-1.3\): \(-1.3=-3.6\sin\left(rac{\pi t}{3} ight)\). Step 2. Divide by \(-3.6\): \(\sin\left(rac{\pi t}{3} ight)=rac{1.3}{3.6}\). Step 3. Take inverse sine: \(rac{\pi t}{3}=\sin^{-1}\left(rac{1.3}{3.6} ight)\). Step 4. Solve for \(t\): \(t=rac{3}{\pi}\sin^{-1}\left(rac{1.3}{3.6} ight)pprox 0.4\). Answer: D
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