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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

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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.

The line of linear function h(x) = –6 + x is not going to enter in the second quadrant so option second will be correct.

What is a graph?

The link between lines and points is described by a graph, which is a diagrammatic representation of a network.

A graph is made up of some points and the distance between two. It doesn't matter how much time the lines are or where the points are located. A node is a name for each constituent in a graph.

Given equation is h(x) = –6 + x

at x = 0 , h(x) = -6 so ( 0 , -6 ) is first point.

at x =1 , h(x) = -5 so ( 1 , -5 ) is second point.

If we draw a line by the two-point which I have attached then you can see it not going to the second quarter.

For more details about the graph

brainly.com/question/16608196

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Ver imagen keshavgandhi04

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