What is the tangent of f(x)=8x^2+8x-5,\at x=-1/2 ?

The tangent of f(x)=8x^2+8x-5,\at x=-1/2 is y=-7

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tangent of f(x)=8x^2+8x-5,\at x=-1/2

tangent of

f (x) = 8 x^{2} + 8 x - 5, at x = - \frac{1}{2}

y = - 7

Solution steps

Find the tangent point:

(- \frac{1}{2}, - 7)

Find the slope of

f (x) = 8 x^{2} + 8 x - 5 : \frac{df ( x )}{d x} = 16 x + 8

Compute the slope of

f (x) = 8 x^{2} + 8 x - 5

x = - \frac{1}{2} : m = 0

Find the line with slope m=

0

and passing through

(- \frac{1}{2}, - 7) : y = - 7

y = - 7

\frac{d}{d x} (\frac{4 x}{5 - tan ( x )})

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

\frac{d}{d x} (x + \frac{4}{x - 4})

\int \frac{1}{sin ( 4 x )} d x

2 x^{2} - 8 x - 7

What is the tangent of f(x)=8x^2+8x-5,\at x=-1/2 ?
The tangent of f(x)=8x^2+8x-5,\at x=-1/2 is y=-7