Question 678378
y = {{{x^2-x-20}}}
0 = (x+4)(x-5)
intercepts at x=-4 and x = 5
complete the square to find the vertex and symmetry
y + 20 = {{{x^2-x}}}
y + 20 + {{{1/4}}} = {{{x^2-x+(1/4)}}}
y + {{{81/4}}} = {{{(x-(1/2))^2}}}
y = {{{(x-(1/2))^2 + (-81/4)}}}
vertex at ({{{1/2}}},{{{-81/4}}}) or (.5,-20.25)
line of symmetry x = {{{1/2}}} or x = .5
...
{{{graph(300,200,-5,7,-25,10,x^2-x-20)}}}
...
...
Using -9x-8y=96, the slope intercept form is...
-9x - 96 = 8y
{{{highlight(y = (-9/8)x - 12)}}}
slope is {{{-9/8}}}
A perpendicular line will have a negative inverse (reciprocal) slope of {{{8/9}}}.
Using the determined slope and the point, plug into the point-slope formula
y - -8 = {{{8/9}}}(x--8)
y+8 = {{{8/9}}}x + {{{8/9}}}(8)
y = {{{(8/9)x + (64/9) - 8(9/9)}}}
{{{highlight_green(y = (8/9)x - 8/9)}}}
...
{{{graph(300,200,-20,5,-10,5,(-9/8)x-12, (8/9)x-(8/9))}}}
.....................
Delighted to help.
-Reading Boosters
Website: www.MyHomeworkAnswers.com 
Wanting for others what we want for ourselves.