What is the tangent of f(x)=-8x^2-7x,\at x=-4 ?

The tangent of f(x)=-8x^2-7x,\at x=-4 is y=57x+128

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tangent of f(x)=-8x^2-7x,\at x=-4

tangent of

f (x) = - 8 x^{2} - 7 x, at x = - 4

y = 57 x + 128

Solution steps

Find the tangent point:

(- 4, - 100)

Find the slope of

f (x) = - 8 x^{2} - 7 x : \frac{df ( x )}{d x} = - 16 x - 7

Compute the slope of

f (x) = - 8 x^{2} - 7 x

x = - 4 : m = 57

Find the line with slope m=

57

and passing through

(- 4, - 100) : y = 57 x + 128

y = 57 x + 128

\int \frac{x ^{2} - 3 x + 12}{( x ^{2} - 4 x + 11 ) ^{2}} d x

\frac{d}{d x} (\frac{1}{x ^{3}} - \frac{2}{x ^{2}})

(x^{4} + 7 x^{2} - 3)^{4}

f (x) = \frac{5}{x ^{2}}, at x = 1

\int \frac{x - 1}{x ^{2} + 3 x + 2} d x

What is the tangent of f(x)=-8x^2-7x,\at x=-4 ?
The tangent of f(x)=-8x^2-7x,\at x=-4 is y=57x+128