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November 28, 2014

November 28, 2014

Posted by **gt** on Tuesday, March 1, 2011 at 6:18pm.

- calculus -
**alx**, Tuesday, March 1, 2011 at 6:53pmy=-2x-3, y=6x-11

- calculus -
**MathMate**, Tuesday, March 1, 2011 at 7:21pmWe are given a fixed point through which the tangents pass.

On the curve (parabola), the slope of the tangent is dy/dx = 2x at the point (x,x²-2).

What we need here is a line that passes through a point (x,f(x)) and (1,-5) with a slope of 2x.

Thus we form the equation:

(y2-y1)/(x2-x1)=slope

(x²-2 - (-5))/(x-1) = 2x

We solve for x=3 or x=-1.

Substitute into equation

(y-y0)=m(x-x0)

to get the formulae alx posted.

- calculus -
**Helper**, Tuesday, March 1, 2011 at 7:52pmy = x^2 - 2

f' = 2x

P(1,-5)

f' = slope m = 2(1) = 2

Equation of the line tangent at P(1,-5)

m = 2

y = mx + b

-5 = 2(1) + b

-5 = 2 + b

b = -7

Equation of the tangent line is,

y = 2x - 7

2x - y = 7

Equation of the normal line at P(1,-5) (The normal line is the line that is perpendicular to the tangent line at the point of tangency).

m = -1/2

y = mx + b

-5 = -1/2(1) + b

-5 = -1/2 + b

b = -10/2 + 1/2

b = -9/2

Equation of the normal line

y = -1/2 x - 9/2

2y = -x - 9

x + 2y = -9

- calculus -
**Helper**, Tuesday, March 1, 2011 at 8:06pmI think I read the question wrong.

I thought the question wanted the slope of the tangent of f(x) at (1,-5) and the two equations at this point.

- calculus -
**Helper**, Tuesday, March 1, 2011 at 8:21pmIgnore my response.

I completely misunderstood the question.(:

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