Find the equation of the line tangent to f(x) = x sqaured. when x = 3

To find the gradient of tangent to a curve at a point we take the first derivative and then substitute the coordinates of the point in.

So this tells us the tangent has gradient 6. Now we know the function so we can work out the coordinates when x=3 since f(3) = 3²=9.

The tangent is a straight line with gradient 6 passing through (3,9) so

review: power rule

use point slope form

where the slope is m and a point on the line is

we can find the point by subsituting 3 for x

so the point is (3,9), ( and )

now we need the slope

do a bit of calculus

take the derivitive of f(x) to get f'(x) which is the slope of f(x) for any value of x

take derivitive, remember the power rule

therefore

f'(x)=2x

find slope at x=3

f'(x)=2x

f'(3)=2(3)

f'(3)=6

so the slope is 6 at that point

so we now know

m=6 and

we can now write that the equation is

if you want it in other forms (I won't show work because if you are asking about derivitives, you should have your algebra down pat)

6x-y=9

y=6x-9

the equation is the ones below (same equation, different forms)

y-9=6(x-3)

6x-y=9

y=6x-9