What is the tangent of f(x)=x^2+8,(-3,17) ?

The tangent of f(x)=x^2+8,(-3,17) is y=-6x-1

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tangent of f(x)=x^2+8,(-3,17)

tangent of

f (x) = x^{2} + 8, (- 3, 17)

y = - 6 x - 1

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

(- 3, 17) : m = - 6

Find the line with slope m=

- 6

and passing through

(- 3, 17) : y = - 6 x - 1

y = - 6 x - 1

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

y = \frac{6 x}{x + 2}

\frac{d}{d x} ((1 + 5 x) (1 - 7 x))

\frac{d}{d x} ((lo g_{e} (x))^{x})

y^{^{''}} + 7 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the tangent of f(x)=x^2+8,(-3,17) ?
The tangent of f(x)=x^2+8,(-3,17) is y=-6x-1