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

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

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tangent of y=x^2+5,(-5,30)

tangent of

y = x^{2} + 5, (- 5, 30)

y = - 10 x - 20

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, 30) : m = - 10

Find the line with slope m=

- 10

and passing through

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

y = - 10 x - 20

2 x

f (x) = (x^{4} - 4 x) e^{x}

- \frac{8}{x ^{9}}

\int (x^{4} + 2 x - 9) ln (x) d x

\frac{3}{x + 1}

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