Thread: Finding normal and tangent to a curve.

1. Finding normal and tangent to a curve.

Hey im doing exam revision and new to this site, the question im stuck on is this"

Find the equation of the tangent and normal for:
f(x)=2x^2-3x+4 at points (3,13)

thanks

2. Differentiate the curve.

Put in x = 3.

Now, find the gradient of the normal so that: gradient of tangent x gradient of normal = -1

You have a point, and a gradient for the tangent line,
a point and a gradient of the normal line.

This should not be too difficult.

3. yeah ive attemped the question and cant get it right im not sure what im doing wrong. can someone please answear it with working out, it will be a huge help

4. The derivative is:

$\displaystyle f'(x) = ....$

The gradient of tangent at x = 3 is:

$\displaystyle f'(3) = ....$

Gradient of normal is then -1/f'(3) = ....

Equation of tangent : y - y1 = f'(3)(x - x1): .....

Equation of normal: y - y1 = (-1/f'(3)) (x - x1) : .....

Moderator edited so that the OP does some work of his/her own

5. Originally Posted by iFuuZe
yeah ive attemped the question and cant get it right im not sure what im doing wrong. can someone please answear it with working out, it will be a huge help
How it works is that you show all your working so that what you're doing wrong can be pointed out to you. Not the other way around.

6. ok i got 4x-3
i put 3 in the rule and got 12-3=9
thats all i can think of

7. What is the equation of the line through the point (a, b) with slope m?

8. Yes, do you know what you just got?

9. Originally Posted by iFuuZe
ok i got 4x-3
i put 3 in the rule and got 12-3=9
thats all i can think of
Yes, f'(3) = 9. Now continue, using post #4 as a guide (you did read post #4, right?)