Thread: 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

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.

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

The derivative is:



The gradient of tangent at x = 3 is:



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

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.

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

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

Yes, do you know what you just got?

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?)