Exercise 121 Page 49

To graph a linear equation, we need to know at least two points through which the line passes. Both equations are written in slope-intercept form. This means we can identify their slope, m, and y-intercept, b which we need to graph them.

y= mx+b Let's identify the slope and y-intercept of each function. c|c y= 3x+15 & y=3 - 3x [0.5em] ⇓ & ⇓ [0.5em] m= 3,b=15 & m= -3,b=3 With this information, we can plot the function. We will start with the second line. It has a y-intercept of 3 and a slope of -3. With this information, we can locate two points on the line and draw its graph.

Using the same procedure, we will plot the second line as well.

The equations intersect at (-2,9).

Let's attempt a graphing solution. Notice that y=2 is a constant function which means it is a horizontal line through the y-coordinate 2.

To draw the graph of the quadratic, we should calculate a few data points through which its parabola passes. |c|c|c| [-1em] x & x^2-3x-8 & y [0.2em] [-1em] -3 & ( -3)^2-3( -3)-8 & 10 [0.2em] [-1em] -2 & ( -2)^2-3( -2)-8 & 2 [0.2em] [-1em] -1 & ( -1)^2-3( -1)-8 & -4 [0.2em] [-1em] 0 & ( 0)^2-3( 0)-8 & -8 [0.2em] [-1em] 1 & ( 1)^2-3( 1)-8 & -10 [0.2em] [-1em] 2 & ( 2)^2-3( 2)-8 & -10 [0.2em] [-1em] 3 & ( 3)^2-3( 3)-8 & -8 [0.2em] Now we can plot the quadratic and mark the points of intersection.

The points intersect at (-2,2) and (5,2).

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Exercise 121 Page 49

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