What is the tangent of f(x)=x^2-9,(-5,16) ?

The tangent of f(x)=x^2-9,(-5,16) is y=-10x-34

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tangent of f(x)=x^2-9,(-5,16)

tangent of

f (x) = x^{2} - 9, (- 5, 16)

y = - 10 x - 34

Solution steps

Find the slope of

f (x) = x^{2} - 9 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 9

(- 5, 16) : m = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 16) : y = - 10 x - 34

y = - 10 x - 34

tan (x) - 1

\int e^{x y} + x y e^{x y} d y

3 y + x^{4} \frac{d y}{d x} = 2 x y

\int \frac{7 x}{1 - x ^{2}} d x

y = cot (2) (cos (x))

What is the tangent of f(x)=x^2-9,(-5,16) ?
The tangent of f(x)=x^2-9,(-5,16) is y=-10x-34