# Question #05248

Feb 6, 2018

In point-slope form: $y - 2 = 4 \left(x - 1\right)$

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

#### Explanation:

We'll begin by finding $f ' \left(x\right)$

Given $f \left(x\right) = {x}^{2} + 2 x - 1$

Then $f ' \left(x\right) = 2 x + 2$

Next we evaluate the derivative at given $x$ value. Since we have the point $\left(1 , 2\right)$, $x = 1$

So

$f ' \left(1\right) = 2 \left(1\right) + 2 = 4$

So the slope of the tangent line is $4$

Now that we have the slope $m = 4$ and a point $\left({x}_{1} , {y}_{1}\right) = \left(1 , 2\right)$ the equation of the tangent line can be found via the point-slope formula:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

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

$y - \left(2\right) = 4 \left(x - 1\right)$

In slope intercept form the equation of the tangent line:

$y = 4 x - 2$

Here's a graph of the function along with its tangent line at $\left(1 , 2\right)$