What is the tangent of y=x^2+3,(-2,7) ?

The tangent of y=x^2+3,(-2,7) is y=-4x-1

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tangent of y=x^2+3,(-2,7)

tangent of

y = x^{2} + 3, (- 2, 7)

y = - 4 x - 1

Solution steps

Find the slope of

y = x^{2} + 3 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} + 3

(- 2, 7) : m = - 4

Find the line with slope m=

- 4

and passing through

(- 2, 7) : y = - 4 x - 1

y = - 4 x - 1

\int_{0}^{\frac{π}{2}} 45 cos^{5} (x) d x

f (x) = 2 x + 4 x^{- 1}

f (x) = 4 x e^{x}

\frac{d}{d x} (x^{3} sin^{2} (5 x))

\int ln (5 x^{2}) d x

What is the tangent of y=x^2+3,(-2,7) ?
The tangent of y=x^2+3,(-2,7) is y=-4x-1