Answer:
[tex]y=8x+50[/tex]
Step-by-step explanation:
We are looking for the same line but with a different y-intercept.
Equation of a line in its Slope–intercept form is given by:
[tex]y=mx+b[/tex]
Where:
[tex]m=slope\\b=y-intercept[/tex]
We know the slope which is the same of the line:
[tex]y=8x-1[/tex]
So:
[tex]m=8[/tex]
And we know a point:
[tex](x_o,y_o)=(-6,2)[/tex]
Let's find the equation of the line parallel to y=8x-1 using the data provided and the next equation:
[tex]y-y_o=m(x-x_o)\\\\y-2=8(x-(-6))\\\\y-2=8x+48\\\\y=8x+50[/tex]
Now let's verify the result.
First, does it contains the point (-6,2)?
Evaluate the function for x=-6:
[tex]y=8(-6)+50\\\\y=-48+50\\\\y=2[/tex]
Second, is it parallel to y=8x-1?
Try to solve the 2x2 system of equations:
[tex]8x-y=1\hspace{3}(1)\\8x-y=-50\hspace{3}(2)[/tex]
Using elimination method:
(1) - (2)
[tex]8x-8x-y-(-y)=1-(-51)\\\\0=-51[/tex]
The system has no solutions. Hence the lines are parallel.
