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

x \to - 9 + lim (\frac{x + 10}{x + 9})

x \to - 2 + lim (- 3 x + 5)

x \to 5 + lim (\frac{∣ x - 5 ∣}{x - 5})

\int \frac{1}{2 + 7 x ^{2}} d x

\frac{d}{d x} ((\frac{1}{2})^{x})

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