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

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- Nov 8th 2010, 11:21 PMiFuuZeFinding 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 PMUnknown008
Differentiate the curve.

Put in x = 3.

This gives your the gradient of the tangent.

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 AMiFuuZe
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 AMUnknown008
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 AMmr fantastic
- Nov 10th 2010, 12:28 AMiFuuZe
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 AMHallsofIvy
What is the equation of the line through the point (a, b) with slope m?

- Nov 10th 2010, 05:13 AMUnknown008
Yes, do you know what you just got?

- Nov 10th 2010, 01:09 PMmr fantastic