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

The tangent of f(x)=2x^2+5x,\at x=-3 is y=-7x-18

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

tangent of

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

y = - 7 x - 18

Solution steps

Find the tangent point:

(- 3, 3)

Find the slope of

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

Compute the slope of

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

x = - 3 : m = - 7

Find the line with slope m=

- 7

and passing through

(- 3, 3) : y = - 7 x - 18

y = - 7 x - 18

\frac{d ^{2} h}{d t ^{2}} = - kh

x \to 2 lim (2 x^{2} - 3 x + 1)

\int \frac{x ^{2}}{4} - 1 d x

y^{^{''}} + 36 y = - 36 sec (6 t)

\frac{d}{d x} (e^{x} + 1)

What is the tangent of f(x)=2x^2+5x,\at x=-3 ?
The tangent of f(x)=2x^2+5x,\at x=-3 is y=-7x-18