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

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

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

tangent of

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

y = - 7 x - 8

Solution steps

Find the tangent point:

(- 2, 6)

Find the slope of

f (x) = 2 x^{2} + x : \frac{df ( x )}{d x} = 4 x + 1

Compute the slope of

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

x = - 2 : m = - 7

Find the line with slope m=

- 7

and passing through

(- 2, 6) : y = - 7 x - 8

y = - 7 x - 8

x \to 0 + lim (ln (x) \cdot ln (1 + x))

4 x^{\frac{5}{4}} + 8 x^{\frac{3}{2}} + 9 x

ln (x - 1)

\frac{d y}{d t} = \frac{e ^{y} + 1}{e ^{y}}

x \to 2 - lim (2 x - 3)

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