What is the tangent of f(x)=110x-16x^2,\at ?

The tangent of f(x)=110x-16x^2,\at is y=(110-32a_{0})x+16a_{0}^2

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tangent of f(x)=110x-16x^2,\at

tangent of

f (x) = 110 x - 16 x^{2}, at

y = (110 - 32 a_{0}) x + 16 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 110 a_{0} - 16 a_{0}^{2})

Find the slope of

f (x) = 110 x - 16 x^{2} : \frac{df ( x )}{d x} = 110 - 32 x

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 = 110 - 32 a_{0}

Find the line with slope m=

110 - 32 a_{0}

and passing through

(a_{0}, 110 a_{0} - 16 a_{0}^{2}) : y = (110 - 32 a_{0}) x + 16 a_{0}^{2}

y = (110 - 32 a_{0}) x + 16 a_{0}^{2}

y = ln e^{1 + x^{2}} + 3 x^{4} + cot^{2} (π x)

x \to 0 - lim (\frac{1}{x} - \frac{1}{x})

(7, - 3), (11, - 4)

\int_{- \infty}^{0} x e^{x^{2}} d x

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

What is the tangent of f(x)=110x-16x^2,\at ?
The tangent of f(x)=110x-16x^2,\at is y=(110-32a_{0})x+16a_{0}^2