What is the tangent of f(x)=x^2+2,(-4,18) ?

The tangent of f(x)=x^2+2,(-4,18) is y=-8x-14

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tangent of f(x)=x^2+2,(-4,18)

tangent of

f (x) = x^{2} + 2, (- 4, 18)

y = - 8 x - 14

Solution steps

Find the slope of

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

Compute the slope of

f (x) = x^{2} + 2

(- 4, 18) : m = - 8

Find the line with slope m=

- 8

and passing through

(- 4, 18) : y = - 8 x - 14

y = - 8 x - 14

\frac{d}{d x} (sec^{2} (x) + csc^{2} (x))

\frac{d}{d x} (g (h, x) (h (x)))

\int \frac{cos ( x )}{7 + sin ^{2} ( x )} d x

f (x) = (x + 2)^{\frac{2}{3}} (2 x^{2} - 5)^{3}

2 - ∣ x ∣, x^{2}

What is the tangent of f(x)=x^2+2,(-4,18) ?
The tangent of f(x)=x^2+2,(-4,18) is y=-8x-14