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

The tangent of f(x)=4x^2+5x,\at x=-2 is y=-11x-16

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

tangent of

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

y = - 11 x - 16

Solution steps

Find the tangent point:

(- 2, 6)

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 = - 2 : m = - 11

Find the line with slope m=

- 11

and passing through

(- 2, 6) : y = - 11 x - 16

y = - 11 x - 16

sin (x), x = \frac{π}{3}

y d x + x (ln (x) - ln (y) - 1) d y = 0, y (1) = e

4 t^{2} - t

x \to 0 lim (\frac{1 - x}{1 - x})

n = 0 \sum \infty \frac{1}{9 ^{n}}

What is the tangent of f(x)=4x^2+5x,\at x=-2 ?
The tangent of f(x)=4x^2+5x,\at x=-2 is y=-11x-16