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What are the coordinates of the point on the graph of f(x)=(x+1)(x+2) at which the tangent is parallel to the line with equation 3x-y-1=0?

a)(-3,2)
b)(-1,0)
c)(0,2)
d)(1,6)

To answer this, I put f(x)=(x+1)(x+2) into the form f(x)=x^2+3x+2 and found the derivative which is f'(x)=2x+3. Next I re-arranged the equation 3x-y-1=0 and got y=3x-1. I don't know what to do after that

2 answers

I found the derivative of y=3x-1 which is y=3. Then I had the two differentiated equations equal one another:

2x+3=3 and solve for x: x=0

Then I subbed x into f(x)=x^2+3x+2 and got y=2

So the coordinate is (0,2)?
good work. Confirm with the graphs at

http://www.wolframalpha.com/input/?i=plot+x%5E2%2B3x%2B2%2C+y%3D3x%2B2+for+-1+%3C%3D+x+%3C%3D+1
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