Find an equation of the normal line to the parabola y = x^2 − 9x + 8 that is parallel to the line x − 7y = 9

Question

my answer was 1/7x+19/2 but it was incorrect, any help is appreciated.

Answer 1

You have the correct slope, because the line we want to be parallel to has a slope of 1/7.

However we need to find the correct ordered pair that our normal line passes through.

To do this we find where the slope is a -7 on our quadratic. So we set the derivative 2x-9 = - 7

Clearly, this happens at x = 1. Why -7 you may ask. Because the tangent slope and normal slope are negative reciprocals of each other and the lines share a point.

The shared x is x =1 which produces a shared y value of 1^2-9(1)+8 = 0

Our normal line must pass through (1,0)

Our normal line is y = 1/7x - 1/7

Answer 2

Given info

a) [perpendicular to]y = x^2 − 9x + 8

n) [parallel to] x − 7y = 9

srep 1

The slope of the given line =1/7

So the line we meed parallal to tis is m=1/7

step 2

Now determine the intersection point (x,y)

The slppe of the oarabola tangent at (x,y)=-1/m=-7

=2x-9

solving for x we get x=1

y=1-9+8=0

so the intersection point is (1,0)

step 3

So the line equation is y=(1/7) [x-1]

or

y=x/7-1/7

Answer 3

The normal must be parallel to y = (x-9)/7, i.e. the normal has slope 1/7.

That means the tangent must have slope -7.

The slope of the parabola is 2x-9

2x-9=-7 => x=1

and the point is (1,0)

The equation of the normal is (1/7) = y/(x-1), i.e. y=(x-1)/7