What is the tangent of f(x)=110x-16x^2,\at x=0 ?

The tangent of f(x)=110x-16x^2,\at x=0 is y=110x

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tangent of f(x)=110x-16x^2,\at x=0

tangent of

f (x) = 110 x - 16 x^{2}, at x = 0

y = 110 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

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

Compute the slope of

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

x = 0 : m = 110

Find the line with slope m=

110

and passing through

(0, 0) : y = 110 x

y = 110 x

\frac{s \cdot ( s + 3 )}{s + 2}

4 t^{3} sin (t)

x y^{^{'}} + y + y^{2} = 0, y (1) = 2

\frac{\partial}{\partial x} (\frac{1}{2} x y^{2})

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

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