What is the tangent of y=x^2+3,(-3,12) ?

The tangent of y=x^2+3,(-3,12) is y=-6x-6

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tangent of y=x^2+3,(-3,12)

tangent of

y = x^{2} + 3, (- 3, 12)

y = - 6 x - 6

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

(- 3, 12) : m = - 6

Find the line with slope m=

- 6

and passing through

(- 3, 12) : y = - 6 x - 6

y = - 6 x - 6

n = 1 \sum \infty \frac{1}{( - 3 ) ^{n}}

\frac{d}{d x} (\frac{2 x ^{2} + 4 x + 8}{x})

\int x (e^{x^{2}}) d x

\frac{d}{d x} (\frac{x}{6} - \frac{6}{x})

\int \frac{x}{x - 1} d x

What is the tangent of y=x^2+3,(-3,12) ?
The tangent of y=x^2+3,(-3,12) is y=-6x-6