1.
Which of the following is NOT a function?
Solution: Step 1. Use the vertical line test. Step 2. A graph represents a function only if every vertical line intersects the graph at most once. Step 3. Option C fails the vertical line test, so it is not a function. Answer: C
2.
Given the graph of \(f(x)\), what is \(f(2)\)?
Solution: Step 1. To find \(f(2)\), look for the \(y\)-value on the graph when \(x=2\). Step 2. From the graph, when \(x=2\), the corresponding \(y\)-value is \(3\). Answer: C
3.
For the equation \(y=-x^2+3\), what is the maximum value of \(y\)?
Solution: Step 1. The equation \(y=-x^2+3\) is a downward-opening parabola because the coefficient of \(x^2\) is negative. Step 2. The vertex is at \((0,3)\). Step 3. Since the parabola opens downward, the maximum \(y\)-value is \(3\). Answer: C
4.
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 graph to identify the highest point in this interval. Step 3. The maximum occurs approximately when \(x=-0.5\). Answer: B
5.
The graph of \(y=\frac{1}{x^2+6x+9}\) is the same as the graph of \(y=\frac{1}{x^2}\) moved left 3 units.
Solution: Step 1. Factor the denominator: \(x^2+6x+9=(x+3)^2\). Step 2. Therefore \(y=rac{1}{x^2+6x+9}=rac{1}{(x+3)^2}\). Step 3. Replacing \(x\) with \(x+3\) moves the graph of \(y=rac{1}{x^2}\) left 3 units. Answer: A
6.
The graph of \(y=(x-3)^7\) is reflected over the \(x\)-axis. Which of the following is the new equation?
Solution: Step 1. A reflection over the \(x\)-axis multiplies all \(y\)-values by \(-1\). Step 2. Starting with \(y=(x-3)^7\), the reflected graph is \(y=-(x-3)^7\). Answer: E
7.
If \(f(x)=3x^2-8x\), then \(f(2a)=12a^2-16a\).
Solution: Step 1. Substitute \(2a\) into \(f(x)=3x^2-8x\). Step 2. \(f(2a)=3(2a)^2-8(2a)\). Step 3. \(f(2a)=3(4a^2)-16a=12a^2-16a\). Answer: B
8.
Which of the graphs shown below represents the base function \(f(x)=x^2\) and the stretched function \(f(x)=-\frac{1}{5}x^2\)?
Solution: Step 1. Compare \(f(x)=x^2\) with \(f(x)=-rac{1}{5}x^2\). Step 2. The negative sign reflects the parabola over the \(x\)-axis. Step 3. Since \(0<rac{1}{5}<1\), the graph is also vertically compressed, making it wider. Step 4. The correct graph is option D. Answer: D
9.
Given the function \(f(x)=x^2-6\) and \(g(x)=2-x\), determine an equation for the combined function \(f(g(x))\).
Solution: Step 1. \(f(g(x))=(g(x))^2-6\). Step 2. Substitute \(g(x)=2-x\): \(f(g(x))=(2-x)^2-6\). Step 3. Expand: \((2-x)^2=x^2-4x+4\). Step 4. Therefore \(f(g(x))=x^2-4x+4-6=x^2-4x-2\). Answer: A
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.
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: C
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: C
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 \(h=2.1\sin\left(\frac{\pi t}{3}\right)\), where \(h\) is the height, in meters, 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 \(1.1\text{ m}\)? Use radian mode.
Solution: Step 1. Set \(h=1.1\): \(1.1=2.1\sin\left(rac{\pi t}{3} ight)\). Step 2. Divide by \(2.1\): \(\sin\left(rac{\pi t}{3} ight)=rac{1.1}{2.1}\). Step 3. Take inverse sine: \(rac{\pi t}{3}=\sin^{-1}\left(rac{1.1}{2.1} ight)\). Step 4. Solve for \(t\): \(t=rac{3}{\pi}\sin^{-1}\left(rac{1.1}{2.1} ight)pprox 0.5\). Answer: A
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