consider the graph of the linear function ( h(x)=-6+\frac{2}{3}x ). which quadrant will the graph not go…

consider the graph of the linear function ( h(x)=-6+\frac{2}{3}x ). which quadrant will the graph not go through and why?\nquadrant i, because the slope is negative and the ( y )-intercept is positive\nquadrant ii, because the slope is positive and the ( y )-intercept is negative\nquadrant iii, because the slope is negative and the ( y )-intercept is positive\nquadrant iv, because the slope is positive and the ( y )-intercept is negative

consider the graph of the linear function ( h(x)=-6+\frac{2}{3}x ). which quadrant will the graph not go through and why?\nquadrant i, because the slope is negative and the ( y )-intercept is positive\nquadrant ii, because the slope is positive and the ( y )-intercept is negative\nquadrant iii, because the slope is negative and the ( y )-intercept is positive\nquadrant iv, because the slope is positive and the ( y )-intercept is negative

Answer

Brief Explanations:

The linear function is in the form (y = mx + b), where (m) is the slope and (b) is the (y)-intercept. For (h(x)=-6+\frac{2}{3}x), (m = \frac{2}{3}>0) (positive slope) and (b=-6<0) (negative (y)-intercept).

  • A positive slope means the line rises from left to right.
  • A negative (y)-intercept (((0, - 6))) means the line crosses the (y)-axis below the origin.
  • In Quadrant II, (x<0) and (y > 0). Since the slope is positive and (y)-intercept is negative, as (x) becomes more negative (moving left from the (y)-intercept), (y=-6+\frac{2}{3}x) will become more negative ((y=-6+\frac{2}{3}x), when (x=-k,k>0), (y=-6-\frac{2}{3}k<0)). So the line does not pass through Quadrant II.

Answer:

Quadrant II, because the slope is positive and the (y)-intercept is negative.