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

The tangent of f(x)=2x^2+3x-2,(-1,-3) is y=-x-4

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

tangent of

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

y = - x - 4

Solution steps

Find the slope of

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

Compute the slope of

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

(- 1, - 3) : m = - 1

Find the line with slope m=

- 1

and passing through

(- 1, - 3) : y = - x - 4

y = - x - 4

\frac{d}{d x} (e^{x^{2} + x y})

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

9 x y^{^{'}} + y = 90 x

g (x) = x^{- \frac{5}{3}} - x + e^{40}

\int \frac{10}{cos ( θ ) - 1} d θ

What is the tangent of f(x)=2x^2+3x-2,(-1,-3) ?
The tangent of f(x)=2x^2+3x-2,(-1,-3) is y=-x-4