What is the equation of the tangent line of `f(x)=3x^2` at `x=1`?

The first objective is to find the slope of the tangent line to the curve of `f(x)=3x^2` at `x=1.` This can be done by finding `f'(1)`, as the derivative at a point represents the slope of the tangent line to the curve at that point (graphically interpreted).

`f'(x)=(3)(2)x^(2-1)`

`f'(x)=6x`

`f'(1)=6` is the slope of the tangent line at `x=1.`

Now we use the point-slope form of a line to find the tangent line equation:

`y-y_0=m(x-x_0)` where `(x_0, y_0)` is a point on the line and `m` is the slope.

We're given `x_0=1,` so `y_0=f(1)=3(1^2)=3`

`m=6,` as calculated above.

`y-3=6(x-1)`

`y-3=6x-6`

`y=6x-3`

We know that the function and it's tangent have the same gradient, so at `f(1)` both will have the same gradient

To get the gradient function of a function we know it is merely the first derivative

`f'(x)=6x`

So now we want to find on our gradient function what the actual gradient will be at one

`f'(1)=6*1=6`

So as the tangent is a straight line the formula for it is `y=mx+c` with m being the gradient

So as the two gradients are equal `6=m`

`y=6x+c`

Now all we have to find out is the y-intercept to do this we have to substiute a co-ordinate from the tangent in. We know that the tangent touches the function when `x=1`

So we can use that co-ordinate pair from the original function

`f(1)=3(1)^2=3`
`(1,3)`

Substitue that into the tangent equation to fine c

`3=6(1)+c` so `-3=c`

`y=6x-3`