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# 4.9: Chapter Test

Difficulty Level: At Grade Created by: CK-12
1. Using the slope-intercept method, graph each of the following functions in the space provided. Show all work needed to draw the graph. State the slope and the y\begin{align*}y-\end{align*}intercept.
1. 3x5y20=0\begin{align*}3x - 5y - 20 = 0\end{align*}
2. 0=2x10\begin{align*}0 = 2x - 10\end{align*}
2. The cost of renting a snowboard at the local ski hill is shown on the graph below.
1. What is the slope of this line?
2. What is the significance of the slope for this problem?
3. What is the y\begin{align*}y-\end{align*}intercept?
4. What does the y\begin{align*}y-\end{align*}intercept represent?
3. The four vertices of quadrilateral ABCD\begin{align*}ABCD\end{align*} are given by A(8,4),B(3,2),C(9,10),\begin{align*}A(-8, 4), B(3, 2), C(9, 10),\end{align*} and D(2,12)\begin{align*}D(-2, 12)\end{align*}. Without graphing, prove that the quadrilateral is a parallelogram.
4. Panther Patty is selling Panther Graduate figurines to the graduates of Glace Bay High. If he sells 4 figurines, he makes a profit of $30. If he sells 10, he makes a profit of$50. His profit and the number of figurines form a linear relationship. He has only 18 figurines to sell.
1. What is the independent variable? What is the dependent variable?
2. Draw a graph that models the relationship.
3. Write an equation to represent this relationship. (Show all work)
4. What is the slope and what does it represent in this problem?
5. What is the profit-intercept and what is its meaning in this problem?
6. Calculate the maximum profit that Panther Patty can make.
7. Write a suitable domain and range for the problem.
5. On the graph draw the line 2x5y+20=0\begin{align*}2x -5y + 20 = 0\end{align*} and the perpendicular line that passes through (3, –2)

(a) 3x5y20=03x5y20=03x3x5y20=03x5y20=3x5y20+20=3x+205y=3x+205y5=35x+2055y5=35x+2045y=35x4\begin{align*}& 3x - 5y - 20 = 0\\ & 3x - 5y - 20 = 0\\ & 3x - 3x -5y - 20 = 0 - 3x\\ & -5y - 20 = -3x\\ & -5y - 20 + 20 = -3x + 20\\ & -5y = -3x + 20\\ & \frac{-5y}{-5} = \frac{-3}{-5}x + \frac{20}{-5}\\ & \frac{\cancel{-5}y}{\cancel{-5}} = \frac{-3}{-5}x + \frac{\overset{-4}{\cancel{20}}}{\cancel{-5}}\\ & \boxed{y = \frac{3}{5}x - 4}\end{align*}

The slope of the line is 35\begin{align*}\frac{3}{5}\end{align*} and the y\begin{align*}y-\end{align*}intercept is (0, –4).

(b) 0=2x1002x=2x2x102x=102x2=1022x2=1052x=5\begin{align*}& 0 = 2x - 10\\ & 0- 2x = 2x -2x -10\\ & -2x = -10\\ & \frac{-2x}{-2} = \frac{-10}{-2}\\ & \frac{\cancel{-2}x}{\cancel{-2}} = \frac{\overset{5}{\cancel{-10}}}{\cancel{-2}}\\ & \boxed{x = 5}\end{align*}

The slope of the line is undefined and there is no y\begin{align*}y-\end{align*}intercept. The line is parallel to the y\begin{align*}y-\end{align*}axis.

(a) (2, 26) (4, 42)

mmmm=y2y1x2x1=422642=162=8\begin{align*}m & = \frac{y_2 - y_1}{x_2 - x_1}\\ m & = \frac{42-26}{4-2}\\ m & = \frac{16}{2}\\ m & = 8\end{align*}

(b) m=Δy()Δx(hr)=81\begin{align*}m = \frac{\Delta y (\)}{\Delta x (hr)} = \frac{8}{1}\end{align*} The significance of the slope is that the cost of renting a snowboard for one hour is eight dollars. (c) The y\begin{align*}y-\end{align*}intercept is (0, 10). (d) The y\begin{align*}y-\end{align*}intercept is the cost intercept. It means that there is an initial fee of10.00 to rent the snowboard.

1. A parallelogram is a quadrilateral that has opposite sides that are parallel. If opposite sides are parallel, then the slopes of the opposite sides must be the same. A(8,4),B(3,2),C(9,10),\begin{align*}A(-8, 4), B(3, 2), C(9, 10),\end{align*} and D(2,12)\begin{align*}D(-2, 12)\end{align*}

mAB=y2y1x2x1 mAB=2438 mAB=243+8 mAB=211 mBC=y2y1x2x1 mBC=10293 mBC=86mBC=43 mCD=y2y1x2x1 mCD=121029 mCD=211mCD=211 mAD=y2y1x2x1 mAD=12428 mAD=1242+8 mAD=86mAD=43\begin{align*}& \ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} && \ m_{BC} = \frac{y_2 - y_1}{x_2 - x_1} && \ m_{CD} = \frac{y_2 - y_1}{x_2 - x_1} && \ m_{AD} = \frac{y_2 - y_1}{x_2 - x_1}\\ & \ m_{AB} = \frac{2-4}{3 - - 8} && \ m_{BC} = \frac{10-2}{9-3} && \ m_{CD} = \frac{12-10}{-2-9} && \ m_{AD} = \frac{12-4}{-2- - 8}\\ & \ m_{AB} = \frac{2-4}{3+8} && \ m_{BC} = \frac{8}{6} && \ m_{CD} = \frac{2}{-11} && \ m_{AD} = \frac{12-4}{-2+8}\\ & \ \boxed{m_{AB} = \frac{-2}{11}} && \boxed{m_{BC} = \frac{4}{3}} && \boxed{m_{CD} = \frac{-2}{11}} && \ m_{AD} = \frac{8}{6}\\ & && && && \boxed{m_{AD} = \frac{4}{3}}\end{align*}

AB||CD\begin{align*}AB || CD\end{align*} and BC||AD\begin{align*}BC || AD\end{align*} Quadrilateral ABCD\begin{align*}ABCD\end{align*} is a parallelogram.

1. (a) The amount of profit Patty makes depends upon the number of figurines he sells. The independent variable is the number of figurines. The dependent variable is the profit. (b) The data is actually discrete data but it has been drawn as {n ε R}\begin{align*}\{ n \ \varepsilon \ R \}\end{align*}. (c) (4, 30) and (10, 50)

mmmmy50505010035010031503100350316.66¯¯¯¯¯16.67=y2y1x2x1=5030104=206=103=mx+b=103(10)+b=1003+b=10031003+b=b=b=b=b=b\begin{align*}m & = \frac{y_2 - y_1}{x_2 - x_1}\\ m & = \frac{50-30}{10-4}\\ m & = \frac{20}{6}\\ m & = \frac{10}{3}\\ \\ y & = mx + b\\ 50 & = \frac{10}{3} (10) + b\\ 50 & = \frac{100}{3} + b\\ 50 - \frac{100}{3} & = \frac{100}{3} - \frac{100}{3} + b\\ 50 - \frac{100}{3} & = b\\ \frac{150}{3} - \frac{100}{3} & = b\\ \frac{50}{3} & = b\\ 16.\overline{66} & = b\\ 16.67 & = b\end{align*}

The equation that models the situation is p=(103)n+16.67\begin{align*}\boxed{p = \left ( \frac{10}{3} \right )n + 16.67}\end{align*}

(c) m=Δp()Δn(#)=103\begin{align*}m = \frac{\Delta p (\)}{\Delta n(\#)} = \frac{10}{3}\end{align*} The slope represents a profit of10.00 for every three figurines that are sold.

(d) The profit intercept is the y\begin{align*}y-\end{align*}intercept which is (0, 16.67). This could represent the money that Panther Patty receives for participating in the fund raiser.

(e) Panther Patty has only 18 figurines to sell. The maximum profit that he could make can be calculated by using the equation of the graph.

\begin{align*}p & = \frac{10}{3}n + 16.67\\ p & = \frac{10}{3} (18) + 16.67\\ p & = \frac{10}{\cancel{3}} \left ( \overset{6}{\cancel{18}} \right ) + 16.67\\ p & = 60 + 16.67\\ p &= \ 76.67\end{align*}

The maximum profit is \$76.67.

(f) domain: \begin{align*}\{ n|0 \le n \le 18, n \ \varepsilon \ R \}\end{align*}

range: \begin{align*}\{p|16.67 \le p \le 76.67, p \ \varepsilon \ R \}\end{align*}

\begin{align*}& 2x - 5y + 20 = 0\\ & 2x - 2x - 5y + 20 = 0 - 2x\\ & -5y + 20 = -2x\\ & -5y + 20 - 20 = -2x - 20\\ & -5y = -2x - 20\\ & \frac{-5y}{-5} = \frac{-2}{-5}x-\frac{20}{-5}\\ & \frac{\cancel{-5}y}{\cancel{-5}} = \frac{-2}{-5}x-\frac{\overset{-4}{\cancel{20}}}{\cancel{-5}}\\ & \boxed{y = \frac{2}{5} x + 4}\end{align*}

The slope of this line is \begin{align*}\frac{2}{5}\end{align*} and the slope of the line perpendicular to this line is \begin{align*}-\frac{5}{2}\end{align*}. The perpendicular line must pass through the point (3, –2).

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