Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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.

Can you answer this question?

1 Answer

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


answered Jul 23, 2013 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App