Question

find an equation of the normal line to the parabola y=x^2-8x+9 at point (3,-6)... help me..

Answer 1

Can you use calculus?

Answer 2

how to use it? i don't know coz it do not give the value of x after differetiate.. so i don't know how to find m..

Answer 3

What do you get for y'(x) in general?

Answer 4

i get 2x -8...

Answer 5

Perfect. This is the equation of the tangents of x^2 - 8x + 9.

Let x = 3 and solve for y'(x) to get the tangent at that point.

Answer 6

More specifically, y'(x) gives you the 'slope' of the tangent at x.

Answer 7

so after that if i get the value of m,, should i substitute the (3,-6) again tu find c?

Answer 8

Let's assume you found the slope at x = 3. This is the slope of the tangent line. The slope of the normal line would be perpendicular to it. What is the relationship of slopes between perpendicular lines?

Answer 9

so, it share the same value of x? right?

Answer 10

Well, the point (3, -6) will be the same. The only difference is, the slope of the normal line will be the negative reciprocal of the slope of the tangent line (since they are perpendicular).