Practice Test (page 391) 1. For each line, count squares on the grid to determine the rise and the run. Use slope = rise

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1 Practice Test (page 91) 1. For each line, count squares on the grid to determine the rise and the. Use slope = rise 4 Slope of AB =, or 6 Slope of CD = 6 9, or Slope of EF = 6, or 4 Slope of GH = 6 4, or The line with slope is EF, which is choice C.. Write the equation x y + = 0 in slope-intercept form. x y + = 0 Subtract x and from each side. y = x Divide each side by. y = x y = x + The graph of this equation has slope. From question 1, the line with slope is CD, which is choice B. a) i) y = x + 5 Since the equation is in slope-intercept form, I use the slope and the y-intercept 5 to graph the line. I mark a point at 5 on the y-axis. I write the slope as, then move squares down and squares right and mark a point. I draw a line through the points. ii) y = 1 (x + ) This equation is in slope-point form. Practice Test Copyright 011 Pearson Canada Inc. 1

2 To identify the coordinates of the point, I write the equation as: y = 1 (x ( )) A point on the line has coordinates (, ), and the slope of the line is 1. I mark a point at (, ), then use the slope 1 to move 1 square up and squares right and mark another point. I draw a line through the points. iii) x 4y 1 = 0 Since the coefficients of x and y in this equation are factors of 1, I can use intercepts to graph the equation. For the x-intercept, substitute y = 0. x 4(0) 1 = 0 x 1 = 0 Solve for x. Add 1 to each side. x = 1 Divide each side by. x = 4 For the y-intercept, substitute x = 0. (0) 4y 1 = 0 4y 1 = 0 Solve for y. Add 1 to each side. 4y = 1 Divide each side by 4. y = I mark a point at 4 on the x-axis and a point at on the y-axis. I draw a line through the points. b) The graph of the equation y = will also have slope. x + 5 has slope, so any line parallel to this line The required line has slope and passes through the point with coordinates (6, ). Substitute: y 1 =, m =, and x 1 = 6 y = (x 6) I know my equation is correct because the slope of the graph of this equation is equal to Practice Test Copyright 011 Pearson Canada Inc.

3 the slope of the given parallel line, and the coordinates of the point on the graph given by this equation are equal to the given coordinates. c) The graph of the equation y = 1 (x + ) has slope 1. The perpendicular line has a slope that is the negative reciprocal of 1 ; that is, its slope is. The required line has slope and passes through the point with coordinates ( 1, ). Substitute: y 1 =, m =, and x 1 = 1 y = (x ( 1)) y = (x + 1) Write this equation in general form. Remove brackets. y = x Add x and to each side. x + y + 1 = 0 This equation is in general form. d) To determine the coordinates of a point P on the graph of x 4y 1 = 0, substitute a value for x, then solve the equation for y. Since the coefficient of y is 4 and the constant term, 1, is a multiple of 4, choose a value of x that is a multiple of 4 so the value of y is an integer. Substitute x = 8 in the equation: x 4y 1 = 0 (8) 4y 1 = 0 Simplify. 4 4y 1 = 0 Collect like terms. 1 4y = 0 Subtract 1 from each side of the equation. 4y = 1 Divide each side by 4. y = The coordinates of P are (8, ). The line passes through P and Q(1, 5). Determine the slope of PQ. change in y Use slope = -coordinates change in x-coordinates Slope = Slope =, or Substitute the coordinates of Q and the slope in the slope-point form of the equation of a line: Substitute: y 1 = 5, m =, and x 1 = 1 y 5 = (x 1) Solve for y. Add 5 to each side. y = (x 1) + 5 Remove brackets. y = x Practice Test Copyright 011 Pearson Canada Inc.

4 y = x y = x + This equation is in slope-intercept form. 4. a) Since I can identify the slope of the line and its y-intercept from the graph, I will write the equation in slope-intercept form: y = mx + b I count squares on the grid to determine the rise and the. Slope = rise 4 Slope =, or From the graph, the y-intercept is. So, an equation is: y = x b) The line is horizontal, so its equation is: y = 1 In general form, this is: y + 1 = 0 c) Since I can calculate the slope and identify the coordinates of a point on the line, I shall use the slope-point form: The coordinates of two points on the line are ( 1, ) and (, 1). I count squares on the grid between these points to determine the rise and the. Slope = rise Slope = 4 Substitute the coordinates of the point (, 1) and the slope 4 So, an equation is: y 1 = (x ) 4 in the equation 5. The cost, C dollars, is a function of the number of people, n, who attend. Use the given information as the coordinates of two points on a sketch of a graph of the function. Plot these points: (600, 11 50) and (400, 650) Practice Test Copyright 011 Pearson Canada Inc. 4

5 The cost per person is the slope of the line through the points on the graph. change in C Slope = -coordinates change in n-coordinates Slope = Slope = Slope = 18 The cost per person is $18. C C 1 = m(n n 1) Substitute: C 1 = 650, m = 18, and n 1 = 400 C 650 = 18(n 400) This is an equation of the function. a), c) To determine the cost for 40 people to attend, substitute n = 40 in the equation: C 650 = 18(n 400) C 650 = 18(40 400) Solve for C. C 650 = 18( 60) C 650 = 1080 C = C = 650 The cost for 40 people to attend is $650. b), c) To determine the number of people who can attend for a cost of $9810, substitute C = 9810 in the equation: C 650 = 18(n 400) = 18(n 400) Solve for n. Remove brackets and simplify. 160 = 18n = 18n n = 50 For a cost of $9810, 50 people can attend. Practice Test Copyright 011 Pearson Canada Inc. 5