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Find the equation of the tangent and normal to the curves. $y=x^{2}-4x-5$ at $\;x=-2$

Note: This is part 1st of a 4 part question, split as 4 separate questions here.

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Toolbox:
  • If $y=f(x)$ then $\large\frac{dy}{dx}$$=f'(x)$ is the rate of change of $y$ w.r.t $x$
  • $\large\frac{dy}{dx_{(x_1,y_1)}}$ is the slope of the tangent to the curve at the point $(x_1,y_1) $ on the curve. It is the slope of the curve at that point.
  • The normal at a point $(x_1,y_1)$ on $y=f(x)$ is perpendicular to the tangent at $(x_1,y_1)$
$y=x^2-4x-5$
Step 1:
When $x=-2,y=(-2)^2-4(-2)-5=4+8-5=7$
We are required to find the equation of the tangent and normal at $(-2,7)$
Step 2:
Slope of the tangent is
$\large\frac{dy}{dx}$$=2x-4$ at $(-2,7)$
=>$m=-4-4=-8$
Equation of the tangent is $(y-7)=-8(x+2)$
$y-7=-8x-16$
$8x+y+9=0$
Step 3:
Normal at $(-2,7)$ is $\perp$ to the tangent
Its slope is $\large\frac{-1}{m}=\large\frac{1}{8}$
Equation of the normal is $(y-7)=\large\frac{1}{8} $$(x+2)$
$8y-56=x+2$
$x-8y+58=0$

 

answered Jul 23, 2013 by meena.p
 

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