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

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Can you use calculus?
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..
What do you get for y'(x) in general?

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i get 2x -8...
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.
More specifically, y'(x) gives you the 'slope' of the tangent at x.
so after that if i get the value of m,, should i substitute the (3,-6) again tu find c?
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?
so, it share the same value of x? right?
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).
The point there is supposed to represent (3, -6). The tangent line is drawn with a slope of m = y'(3), and the normal line is also drawn perpendicular to it.
ok i understand now...thanks for your explanation....
y'(3) = 2(3) - 8 = -2 = slope of the tangent line Thus the slope of the normal line would be 1/2 and pass through (3, -6).
but i little bit confuse,, how do u know the shape of graph was perpendicular?
Do you mean the direction of the normal line? A normal line is just a line perpendicular to whatever shape we're talking about. More specifically, a normal line is perpendicular to a tangent line.
ok i understand..thank u,, :)

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