Please, could you proof read your input before posting a question? The character " â " does not belong here. "x2" should be written as x^2, where the "^" denotes exponentiation.
I'm taking the liberty of correcting your input: "Find an equation of the normal line to the parabola y = x2 â 9x + 7 that is parallel to the line x â 7y = 2." may be:
Find an equation of the normal line to the parabola y = x^2 - 9x + 7 that is parallel to the line x - 7y = 2.
How do you find the slope of a normal line to the graph of a given quadratic? Differentiate y = x^2 - 9x + 7: dy/dx = 2x - 9
You can see that this slope varies with x. You have not named an x-value here, so I will have to stop at dy/dx = 2x - 9.
The slope of a line perpendicular to the tangent line is the negative reciprocal of the slope of the tangent line, or
-1
------- which obviously depends upon the x value you choose or are given.
(2x)
Now focus on the given line x - 7y = 2. Solving for y, 7y =x-2. The slope of this line is (1/7). Thus,
-1 1
--------- = ---------- Cross-multiplying, -7 = 2x; x = -3.5
2x 7
So it appears we DO have the x-value at the point of tangency on the curve.
Find the associated y-value by subst. x = -3.5 into the eq'n of the parabola.
Call it Y.
Then the eq'n of the normal line to the parabola that is parallel to x - 7y = 2
is y-Y = (1/7)(x-[-3.5])
