What is the tangent of f(x)=x^2+4,(-3,13) ?

The tangent of f(x)=x^2+4,(-3,13) is y=-6x-5

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tangent of f(x)=x^2+4,(-3,13)

tangent of

f (x) = x^{2} + 4, (- 3, 13)

y = - 6 x - 5

Solution steps

Find the slope of

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

Compute the slope of

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

(- 3, 13) : m = - 6

Find the line with slope m=

- 6

and passing through

(- 3, 13) : y = - 6 x - 5

y = - 6 x - 5

\int \frac{cos ^{3} ( 3 t )}{sin ^{5} ( 3 t )} d t

y^{^{'}} + \frac{2 y}{t - 2} = \frac{sin ( t )}{t - 2}

\frac{\partial}{\partial x} (3 x y + 4 x z + 5 yz - 3)

\frac{d}{d x} (3 cot (x))

\int 2 - 2 cos (t) d t

What is the tangent of f(x)=x^2+4,(-3,13) ?
The tangent of f(x)=x^2+4,(-3,13) is y=-6x-5