Question

Question #05248

Answer 1

In point-slope form: `y-2=4(x-1)`

In slope intercept form: `y=4x-2`

Explanation:

We'll begin by finding `f'(x)`

Given `f(x)=x^2+2x-1`

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

Next we evaluate the derivative at given `x` value. Since we have the point `(1,2)`, `x=1`

So

`f'(1)=2(1)+2=4`

So the slope of the tangent line is `4`

Now that we have the slope `m=4` and a point `(x_1,y_1)=(1,2)` the equation of the tangent line can be found via the point-slope formula:

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

So we substitute the known values and we'll get the equation of the tangent line in point-slope form:

`y-(2)=4(x-1)`

In slope intercept form the equation of the tangent line:

`y=4x-2`

Here's a graph of the function along with its tangent line at `(1,2)`