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

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

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

tangent of

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

y = - 10 x - 20

Solution steps

Find the slope of

f (x) = x^{2} + 5 : \frac{df ( x )}{d x} = 2 x

Compute the slope of

f (x) = 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

x y^{^{''}} - 2 y^{^{'}} + 4 = 0

(- 3.7) (- 4.12)

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

x \to 100 lim (20 + 0.188 x)

\int y 6 + 12 y - 36 y^{2} d y

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