For this case we have the following data:
Slope, [tex]m = \frac{2}{3}[/tex]
[tex](x, y) = (- 6,4)[/tex]
By definition, the equation of the slope-intercept form is given by:
[tex]y = mx + b[/tex]
We must find the cut point b, for this we substitute the given point and the slope in the equation:
[tex]4 =\frac{2}{3}(-6) + b[/tex]
[tex]4 =-\frac{12}{3}+ b[/tex]
[tex]4 = -4 + b\\4 + 4 = b\\b = 8[/tex]
Thus, the equation is given by:
[tex]y =\frac{2}{3}x + 8[/tex]
Now, we look for the points of intersection with the x and y axes respectively:
Point of intersection with the y axis:
We do[tex]x = 0[/tex] and substitute in the equation found:
[tex]y =\frac{2}{3}(0) +8\\y = 8[/tex]
Thus, the point of intersection with the y-axis is (0,8).
Point of intersection with the x axis:
We make [tex]y = 0[/tex]and substitute in the equation found:
[tex]0 =\frac{2}{3}x+ 8[/tex]
Clear x:
[tex]\frac{2}{3}x= -8\\ 2x = -8 * 3[/tex]
[tex]x = -\frac{24}{2}\\x = -12[/tex]
Thus, the point of intersection with the x-axis is (-12.0).
Answer:
[tex]y =\frac{2}{3}x + 8[/tex]
See attached image
