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

The tangent of f(x)=2(x-4)^2 is y=4(a_{0}-4)x-2a_{0}^2+32

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tangent of f(x)=2(x-4)^2

tangent of

f (x) = 2 (x - 4)^{2}

y = 4 (a_{0} - 4) x - 2 a_{0}^{2} + 32

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 (a_{0} - 4)^{2})

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 4 (a_{0} - 4)

Find the line with slope m=

4 (a_{0} - 4)

and passing through

(a_{0}, 2 (a_{0} - 4)^{2}) : y = 4 (a_{0} - 4) x - 2 a_{0}^{2} + 32

y = 4 (a_{0} - 4) x - 2 a_{0}^{2} + 32

f (x) = 10 x - 3 x^{2}

\int \frac{x}{( 16 - x ^{2} ) ^{2}} d x

y = \frac{- 4 x}{x ^{2} + 1}, (1, - 2)

\int \frac{10}{x + 10 + x} d x

7 t \frac{d y}{d t} + y = 0

What is the tangent of f(x)=2(x-4)^2 ?
The tangent of f(x)=2(x-4)^2 is y=4(a_{0}-4)x-2a_{0}^2+32