What is the tangent of f(x)=x^2-8,(-5,17) ?

The tangent of f(x)=x^2-8,(-5,17) is y=-10x-33

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tangent of f(x)=x^2-8,(-5,17)

tangent of

f (x) = x^{2} - 8, (- 5, 17)

y = - 10 x - 33

Solution steps

Find the slope of

f (x) = x^{2} - 8 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = x^{2} - 8

(- 5, 17) : m = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 17) : y = - 10 x - 33

y = - 10 x - 33

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

x \to - 3 lim (- 6)

(- 1.2) (- 5.2)

\int - 17 sin^{4} (x) cos (x) d x

x^{3} \frac{d y}{d x} = x^{2} y - y^{3}

What is the tangent of f(x)=x^2-8,(-5,17) ?
The tangent of f(x)=x^2-8,(-5,17) is y=-10x-33