What is the tangent of f(x)=4x^2+5x,\at x=-3 ?

The tangent of f(x)=4x^2+5x,\at x=-3 is y=-19x-36

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tangent of f(x)=4x^2+5x,\at x=-3

tangent of

f (x) = 4 x^{2} + 5 x, at x = - 3

y = - 19 x - 36

Solution steps

Find the tangent point:

(- 3, 21)

Find the slope of

f (x) = 4 x^{2} + 5 x : \frac{df ( x )}{d x} = 8 x + 5

Compute the slope of

f (x) = 4 x^{2} + 5 x

x = - 3 : m = - 19

Find the line with slope m=

- 19

and passing through

(- 3, 21) : y = - 19 x - 36

y = - 19 x - 36

\int \frac{4}{- x ^{2} + 6 x + 5} d x

arcsec (x) \cdot arccos (x)

\int - y^{2} d y

\frac{d}{d x} (\frac{- 1}{4 x ^{\frac{3}{2}}})

What is the tangent of f(x)=4x^2+5x,\at x=-3 ?
The tangent of f(x)=4x^2+5x,\at x=-3 is y=-19x-36