What is the tangent of y=x^2-5,(-5,20) ?

The tangent of y=x^2-5,(-5,20) is y=-10x-30

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tangent of y=x^2-5,(-5,20)

tangent of

y = x^{2} - 5, (- 5, 20)

y = - 10 x - 30

Solution steps

Find the slope of

y = x^{2} - 5 : \frac{d y}{d x} = 2 x

Compute the slope of

y = x^{2} - 5

(- 5, 20) : m = - 10

Find the line with slope m=

- 10

and passing through

(- 5, 20) : y = - 10 x - 30

y = - 10 x - 30

f (x) = \frac{1}{t ^{2}}

\int 43 tan^{5} (x) sec^{4} (x) d x

y = - x^{2} + 6 x, y = x^{2} - 4 x

\frac{4}{( 11 x - 3 ) ^{9}}

\int_{0}^{\frac{π}{2}} cos (x) d x

What is the tangent of y=x^2-5,(-5,20) ?
The tangent of y=x^2-5,(-5,20) is y=-10x-30