What is the tangent of y=x^2+2,(-3,11) ?

The tangent of y=x^2+2,(-3,11) is y=-6x-7

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tangent of y=x^2+2,(-3,11)

tangent of

y = x^{2} + 2, (- 3, 11)

y = - 6 x - 7

Solution steps

Find the slope of

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

Compute the slope of

y = x^{2} + 2

(- 3, 11) : m = - 6

Find the line with slope m=

- 6

and passing through

(- 3, 11) : y = - 6 x - 7

y = - 6 x - 7

x \to 1 lim (\frac{( 26 - x - 5 )}{17 - x - 4})

y^{^{''}} - 2 y^{^{'}} - 3 y = 4 e^{3 t}

\int - 2 e^{c o s (x)} sin (x) d x

f (x) = 21

\int cos^{3} (4 x) sin^{2} (4 x) d x

What is the tangent of y=x^2+2,(-3,11) ?
The tangent of y=x^2+2,(-3,11) is y=-6x-7