Question

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

Answer 1

review: power rule

`(d)/(dx) x^n=nx^(n-1)`

use point slope form

`y-y_1=m(x-x_1)`

where the slope is m and a point on the line is
`(x_1, y_1)`

we can find the point by subsituting 3 for x

`f(x)=x^2`

`f(3)=(3)^2`

`f(3)=9`

so the point is (3,9), (
`x_1=3` and
`y_1=9`)

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

`f(x)=x^2`

take derivitive, remember the power rule

`(d)/(dx) x^2=2x^(2-1)=2x^1=2x`

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
`(x_1,y_1)=(3,9)`

we can now write that the equation is

`y-9=6(x-3)`

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

Answer 2

`f(x)=x^2`

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.

`f'(x) = 2x`

`f'(3) = 6`

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

`y-y_0=m(x-x_0)`

`y-9=6(x-3)`

`y=6x-9`