Question

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

Answer 1

`y=8x-9`

Explanation:

First, find the point of tangency, which is the point on the function which the tangent line will intercept:

`f(3)=3^2+2(3)=15`

Thus, the tangent line passes through the point `(3,15)`.

To find the slope of the tangent line, find the value of the derivative at `x=3`.

To find the derivative of the function, use the power rule.

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

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

The slope of the tangent line is

`f'(3)=2(3)+3=8`

So, we know that the tangent line passes through the point `(3,15)` and has a slope of `8`.

These can be related as a linear equation in point-slope form, which takes a point `(x_1,y_1)` and slope `m`:

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

Thus, the equation of the tangent line is

`y-15=8(x-3)`

Which can be rewritten as

`y=8x-9`

Graphed are the original function and its tangent line:

graph{(x^2+2x-y)(y-8x+9)=0 [-2, 7, -11, 40]}