Question #05248

1 Answer
Feb 6, 2018

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)#

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