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Show that the point........?

+2 votes
(-1,5) is on the graph of y=5x^(2), and find the equation of the tangent line at the point (-1,5).
asked Feb 1, 2013 in CALCULUS by linda Scholar

1 Answer

+2 votes

Given that point (-1, 5) and the graph equation is y = 5x2

y = 5x2

Apply derivative each side.

y' = (5x2)'

The derivative of xn = nxn-1

= 5(2x) = 10x.

The slope of the tangent line must equal the derivative of y at x  m = y'(x)

There fore slope m = y' = 10x

But x = -1, m = y' = 10(-1) = -10.

Recall "point-slope" formula to write the equation is (y - y₁) = m(x -x₁)

(y - 5) = (-10)[x - (-1)]

(y - 5) = -10(x +1)

(y - 5) = -10x - 10

Add 5 to each side.

y = -10x - 5

There fore the tangent line equation is y = -10x - 5.

answered Feb 1, 2013 by richardson Scholar
Thank you!!!

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