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

The tangent of f(x)=2x^2+3x,\at x=-2 is y=-5x-8

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

tangent of

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

y = - 5 x - 8

Solution steps

Find the tangent point:

(- 2, 2)

Find the slope of

f (x) = 2 x^{2} + 3 x : \frac{df ( x )}{d x} = 4 x + 3

Compute the slope of

f (x) = 2 x^{2} + 3 x

x = - 2 : m = - 5

Find the line with slope m=

- 5

and passing through

(- 2, 2) : y = - 5 x - 8

y = - 5 x - 8

f (x) = e^{3 e^{x}}

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

\frac{x}{cos ( x )}

x^{3} - 6 x^{2} + 9 x + 2

x^{2 c o s (x)}

What is the tangent of f(x)=2x^2+3x,\at x=-2 ?
The tangent of f(x)=2x^2+3x,\at x=-2 is y=-5x-8