Question #05248

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Answer 1 · 2018-02-06T03:08:59

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

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

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 $(1,2)$

enter image source here