Mathematics 40 Online OpenStudy (anonymous):

Find the slope of the tangent line to the parabola y = 4 x^2 + 2 x + 7 at x0 = -2 OpenStudy (anonymous):

(The tangent line to a curve at a point is the line passing through that point whose slope is the same as the slope of the curve at that point.) OpenStudy (anonymous):

The equation of this tangent line can be written in the form y - y0 = m(x - x0) where y0 is: ?? OpenStudy (anonymous):

the slope of the tangent line is the first derivative of the parabola which is 8x+2, so m= 8 OpenStudy (anonymous):

yes, I figured out that the tangent line is: 8x + 2, which I obtained by taking the derivative of the parabola. Plugging in x0 = -2 into 8x+2 we get -14, which is the answer to the first part. I just don't know what y0 equals in y - y0 = m(x - x0). I suspect this to be correct: y - y0 = 8(x+2), but I don't know how to proceed. Can I just plug in arbitrary values of x and y to get y0? OpenStudy (anonymous):

Yes, the slope of the tangent line is m = f ' (-2) = -14 as you found. y0 is just notation representing f(x0). So y0 = f(-2) = 19 for your problem. The tangent line is then y = 19 -14(x+2). Note that I moved the y0 over to the right side of the equation (personal preference on my part....). OpenStudy (anonymous):

Ok, thanks!!! :)

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