# Finding normal and tangent to a curve.

• Nov 8th 2010, 11:21 PM
iFuuZe
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
• Nov 8th 2010, 11:46 PM
Unknown008
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.
• Nov 9th 2010, 12:12 AM
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
• Nov 9th 2010, 12:25 AM
Unknown008
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
• Nov 9th 2010, 03:10 AM
mr fantastic
Quote:

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.
• Nov 10th 2010, 12:28 AM
iFuuZe
ok i got 4x-3
i put 3 in the rule and got 12-3=9
thats all i can think of
• Nov 10th 2010, 04:01 AM
HallsofIvy
What is the equation of the line through the point (a, b) with slope m?
• Nov 10th 2010, 05:13 AM
Unknown008
Yes, do you know what you just got?
• Nov 10th 2010, 01:09 PM
mr fantastic
Quote:

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?)