Question

How do you find the equation of a line tangent to the function y=−3x2−25 at (-4, 1/3)?

Answer 1

let f(x)= y = -3x^2-25

f'(x) = -6x-0 (differentiate the function)

we have to find slope of tangent at x= -4

in f'(-4) = -6 × -4 = 24 (slope)

equation:

y - y1 = m(x - x1) ( m is slope of tangent)

y - 1/3 = 24(x- (-4)) (point slope form)

y - 1/3 = 24x +96

24x - y + 289/3 = 0 (required equation)

Hope this helps you mate.

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