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3.11: Chapter Test

Difficulty Level: At Grade Created by: CK-12
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  1. \begin{align*}f(x) = 4x-8\end{align*}f(x)=4x8 is a function.
    1. Find \begin{align*}f(-3)\end{align*}f(3)
    2. Find ‘\begin{align*}x\end{align*}x’ if \begin{align*}f(x) = 28\end{align*}f(x)=28
    3. Is the following graph a function? Why or why not?
    4. What are the domain and range of the above graph?
  2. What are the coordinates of the points plotted on the following Cartesian plane?
  3. The cost of renting the ice surface at the local arena is shown in the table below.
Time (hr) 0 1 2 3 4
Cost ($) $200 $250 $300 $350 $400

(a) Write a linear function to represent the cost of renting the ice surface.

(b) Draw a graph to represent the problem. Label the axis to match the problem.

(c) Write a suitable domain and range for the problem.

    1. If the graph of \begin{align*}y=x^2\end{align*}y=x2 undergoes a vertical stretch of \begin{align*}\frac{2}{3}\end{align*}23, what will be the \begin{align*}y-\end{align*}yvalues for the graph?
    2. Write a mapping rule for the equation \begin{align*}-\frac{1}{2} (y-6) = (x+7)^2\end{align*}12(y6)=(x+7)2.
    3. For the above mapping rule, create a table of values.
  1. Answer the following questions with respect to this graph.
    1. What is the vertex of the graph? ____________________
    2. Does the graph have a max value or a min. value? __________Of what? __________
    3. What is the equation of the axis of symmetry? ____________________
    4. What is the domain of the graph? ____________________
    5. What is the range of the graph? ____________________
  2. For the function \begin{align*}6x+3y-24=0\end{align*}6x+3y24=0, determine the \begin{align*}x\end{align*}x and \begin{align*}y-\end{align*}yintercepts algebraically. Use the intercepts to draw the graph.
  3. For the following graph, list the transformations of \begin{align*}y=x^2\end{align*}y=x2 and write the equation to describe the graph.

Answers to Test

  1. \begin{align*}f(x)=4x-8\end{align*}f(x)=4x8

(a) \begin{align*}& \quad f(x) = 4x - 8\\ & \ f(-3) = 4(-3) - 8\\ & \ f(-3) = -12-8\\ & \boxed{f(-3) = -20}\end{align*}f(x)=4x8 f(3)=4(3)8 f(3)=128f(3)=20

(b) \begin{align*}& \ \ f(x) = 4x - 8\\ & \quad \ \ 28 = 4x - 8\\ & 28 + 8 = 4x - 8 + 8\\ & \quad \ \ 36 = 4x\\ & \quad \ \frac{36}{4} = \frac{4x}{4}\\ & \quad \ \frac{\overset{9}{\cancel{36}}}{\cancel{4}} = \frac{\cancel{4}x}{\cancel{4}}\\ & \quad \ \ \boxed{9 = x}\end{align*}  f(x)=4x8  28=4x828+8=4x8+8  36=4x 364=4x4 3694=4x4  9=x

(c) The graph is not a function. A vertical line drawn through the graph will intersect it in more than one place.

(d) domain: \begin{align*}\{ x | -3 \le x \le 7, x \ \varepsilon R \}\end{align*}{x|3x7,x εR}

range: \begin{align*}\{ y | -2 \le y \le 8, y \ \varepsilon \ R\end{align*}{y|2y8,y ε R

  1. The coordinates of the plotted points are: \begin{align*}A (3,4); B (2,-4); C (-1,-3); D (-2,5); E (-6,1)\end{align*}A(3,4);B(2,4);C(1,3);D(2,5);E(6,1)
    1. The cost of renting the ice surface is modeled by the function \begin{align*}\boxed{c=50t+200}\end{align*}c=50t+200 where ‘\begin{align*}c\end{align*}c’ is the cost in dollars and ‘\begin{align*}t\end{align*}t’ is the time in hours.
    2. domain: \begin{align*}\{t | t \ge 0, t \ \varepsilon \ R \}\end{align*}{t|t0,t ε R}

    range: \begin{align*}\{c | c \ge 200, c \ \varepsilon \ R \}\end{align*}{c|c200,c ε R}

    1. The \begin{align*}y-\end{align*}yvalues are 1, 4, 9. If the graph undergoes a vertical stretch of \begin{align*}\frac{2}{3}\end{align*}23, the new \begin{align*}y-\end{align*}yvalues will be \begin{align*}\frac{2}{3}(1) = \frac{2}{3}; \frac{2}{3}(4) = \frac{8}{3} = 2 \frac{2}{3}; \frac{2}{3}(9) =6\end{align*}23(1)=23;23(4)=83=223;23(9)=6.
    2. \begin{align*}-\frac{1}{2} (y-6) = (x+7)^2\end{align*}12(y6)=(x+7)2 Mapping Rule: \begin{align*}\boxed{(x,y) \rightarrow (x-7,-2y+6)}\end{align*}(x,y)(x7,2y+6)
    3. Table of Values:
\begin{align*}x \rightarrow x - 7\end{align*}xx7 \begin{align*}y \rightarrow -2y + 6\end{align*}y2y+6
\begin{align*}-3\end{align*}3 –10 9 –12
–2 –9 4 –2
–1 –8 1 4
0 –7 0 6
1 –6 1 4
2 –5 4 –2
3 –4 9 –12
    1. Vertex (-4, 3)
    2. a maximum value of 3
    3. \begin{align*}x = -4\end{align*}x=4
    4. \begin{align*}D = \{ x | x \ \varepsilon \ R \}\end{align*}D={x|x ε R}
    5. \begin{align*}R = \{ y | y \le 3, y \ \varepsilon \ R \}\end{align*}R={y|y3,y ε R}
  1. The transformations of \begin{align*}y = x^2\end{align*}y=x2 are: \begin{align*}VR \rightarrow NO; \ VS \rightarrow \frac{1}{2}; \ VT \rightarrow -5; \ HT \rightarrow +3\end{align*}VRNO; VS12; VT5; HT+3 The equation to model the graph is \begin{align*}\boxed{2(y+5) = (x-3)^2}\end{align*}2(y+5)=(x3)2

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Jan 16, 2013
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