what is the slope intercept form of the linear equation with a graph that passes through (-4,7) and (6,-8)

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gmany gmany · Answer 1 · 2018-11-14T22:09:53+01:00

The slope-intercept form:

[tex]y=mx+b[/tex]

m - slope

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

b - y-intercept.

We have the points (-4, 7) and (6, -8). Substitute:

[tex]m=\dfrac{-8-7}{6-(-4)}=\dfrac{-15}{10}=-\dfrac{3}{2}[/tex]

Then, we have

[tex]y=-\dfrac{3}{2}x+b[/tex]

Put the coordinates of the point (-4, 7) to the equation of line:

[tex]7=-\dfrac{3}{2}(-4)+b\\\\7=3(2)+b\\\\7=6+b\quuad\text{subtract 6 from both sides}\\\\1=b[/tex]

Answer:

[tex]y=-\dfrac{3}{2}x+1[/tex]

CharlieBrown2002 CharlieBrown2002 · Answer 2 · 2018-11-14T22:15:41+01:00

Explanation:

Slope-intercept form: → [tex]y=mx+b[/tex]

m: represents the slope and is constant.

b: represents the y-intercept.

The y-intercept is the point on a graph at which the graph crosses the y-axis.

You had to used rise/run.

[tex]\frac{rise}{run}[/tex]

[tex]m=\frac{rise}{run}[/tex]

[tex]Slope=\frac{y^2-y^1}{x^2-x^1}[/tex]

[tex]rise=y^2-y^1[/tex]

[tex]run=x^2-x^1[/tex]

[tex](x^1,y^1)=(-4,6)[/tex]

[tex](x^2,y^2)=(7,-8)[/tex]

[tex]rise=y^2-y^1=-8-7=-15[/tex]

[tex]run=x^2-x^1=6--4=10[/tex]

[tex]-15/10[/tex]

But the slope is -15.

But the y-intercept is 10.

Hope this helps!