What is the tangent of f(x)=2x^2+5x-1,\at x=-3 ?

The tangent of f(x)=2x^2+5x-1,\at x=-3 is y=-7x-19

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tangent of f(x)=2x^2+5x-1,\at x=-3

tangent of

f (x) = 2 x^{2} + 5 x - 1, at x = - 3

y = - 7 x - 19

Solution steps

Find the tangent point:

(- 3, 2)

Find the slope of

f (x) = 2 x^{2} + 5 x - 1 : \frac{df ( x )}{d x} = 4 x + 5

Compute the slope of

f (x) = 2 x^{2} + 5 x - 1

x = - 3 : m = - 7

Find the line with slope m=

- 7

and passing through

(- 3, 2) : y = - 7 x - 19

y = - 7 x - 19

\frac{d}{d x} ((4 x^{2} + 5) (5 x + 2)^{4})

\frac{d}{d t} ((t - 1)^{2})

\frac{d}{d x} (3^{c o s (x)})

\int \frac{5}{7 + 5 x} d x

\frac{d}{d x} (x^{2} + A x + B)

What is the tangent of f(x)=2x^2+5x-1,\at x=-3 ?
The tangent of f(x)=2x^2+5x-1,\at x=-3 is y=-7x-19