What is the tangent of f(x)=8-2x^2,\at x=0 ?

The tangent of f(x)=8-2x^2,\at x=0 is y=8

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tangent of f(x)=8-2x^2,\at x=0

tangent of

f (x) = 8 - 2 x^{2}, at x = 0

y = 8

Solution steps

Find the tangent point:

(0, 8)

Find the slope of

f (x) = 8 - 2 x^{2} : \frac{df ( x )}{d x} = - 4 x

Compute the slope of

f (x) = 8 - 2 x^{2}

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, 8) : y = 8

y = 8

\int x 2 x + 5 d x

f (x) = 25 x^{\frac{4}{5}}

\int tan^{2} (x) + tan^{4} (x) d x

n = 1 \sum \infty (- 5)^{n} x^{n}

What is the tangent of f(x)=8-2x^2,\at x=0 ?
The tangent of f(x)=8-2x^2,\at x=0 is y=8