# Find the equation of the tangent line to the function `f(x)=x^2` at x=3

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### 1 Answer

First, solve for the value of y when x=3 to get the point of tangency.

`f(x)= x^2=3^2=9`

So, the point of tangency is (3,9).

Next, determine the slope of the line tangent to the given function at (3,9). To do so, take the derivative of f(x).

`f'(x)= (x^2)'`

`f'(x)=2x`

Then, plug-in the value of x.

`f'(3)=2*3=6`

Hence, the slope of the tangent line is 6.

Since the tangent line contains the point (3,9), use the point-slope form to determine its equation.

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

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

Then, distribute 6 to x-3.

`y-9=6x-18`

And, add 9 on both side of the equation to get y only.

`y-9+9=6x-18+9`

`y=6x-9` **Hence, the equation of the tangent line is `y=6x-9` .**