Question

Consider the graph of the linear function h(x) = –6 + x. Which quadrant will the graph not go through and why? Quadrant I, because the slope is negative and the y-intercept is positive Quadrant II, because the slope is positive and the y-intercept is negative Quadrant III, because the slope is negative and the y-intercept is positive Quadrant IV, because the slope is positive and the y-intercept is negative

Answer 1

Quadrant II, because the slope is positive and the y intercept is negative

Answer 2

You'd benefit from graphing h(x) = –6 + x.  To do this, obtain the slope and y-intercept of this straight line from h(x) = –6 + x:

Slope is +1 = 1/1 = rise / run

y-intercept is -6, that is, (0,-6)

To graph this, plot the point (0,-6).  Starting at this point, go 1 unit to the right, to (1,-6), and then from (1,-6) go 1 unit up, to (1,-5).  Draw a line thru (0,-6) and (1,-5).  You will see that this line is BELOW the origin; it never enters Quadrant II.