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

The tangent of f(x)=2x^2-7,\at x=-2 is y=-8x-15

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tangent of f(x)=2x^2-7,\at x=-2

tangent of

f (x) = 2 x^{2} - 7, at x = - 2

y = - 8 x - 15

Solution steps

Find the tangent point:

(- 2, 1)

Find the slope of

f (x) = 2 x^{2} - 7 : \frac{df ( x )}{d x} = 4 x

Compute the slope of

f (x) = 2 x^{2} - 7

x = - 2 : m = - 8

Find the line with slope m=

- 8

and passing through

(- 2, 1) : y = - 8 x - 15

y = - 8 x - 15

4 x + y^{2} = 13, x = 3 y

(ln (3 x))^{^{'}}

6 e^{- 5 t} + 5 t

\frac{\partial}{\partial x} ((x^{2} + y) (x - y^{2}))

\int x^{31} sin (x^{32}) d x

What is the tangent of f(x)=2x^2-7,\at x=-2 ?
The tangent of f(x)=2x^2-7,\at x=-2 is y=-8x-15