Browse Questions

# 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.

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)$
$8y-56=x+2$
$x-8y+58=0$