What is the tangent of f(x)=2-2x^2,-16,\at x=3 ?

The tangent of f(x)=2-2x^2,-16,\at x=3 is y=-12x+20

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tangent of f(x)=2-2x^2,-16,\at x=3

tangent of

f (x) = 2 - 2 x^{2}, - 16, at x = 3

y = - 12 x + 20

Solution steps

Find the tangent point:

(3, - 16)

Find the slope of

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

Compute the slope of

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

x = 3 : m = - 12

Find the line with slope m=

- 12

and passing through

(3, - 16) : y = - 12 x + 20

y = - 12 x + 20

x \to - 4 lim (10 x^{2} + 34 x - 24)

n = 1 \sum \infty 2 - 5 (0.9)^{n}

n = 0 \sum \infty \frac{1}{n} - \frac{1}{n + 1}

x \to - 5 - lim (\frac{x + 2}{x + 5})

\frac{3}{s + 4}

What is the tangent of f(x)=2-2x^2,-16,\at x=3 ?
The tangent of f(x)=2-2x^2,-16,\at x=3 is y=-12x+20