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.
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