What is the tangent of f(x)=2x^2-18x+5,\at x=0 ?

The tangent of f(x)=2x^2-18x+5,\at x=0 is y=-18x+5

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tangent of f(x)=2x^2-18x+5,\at x=0

tangent of

f (x) = 2 x^{2} - 18 x + 5, at x = 0

y = - 18 x + 5

Solution steps

Find the tangent point:

(0, 5)

Find the slope of

f (x) = 2 x^{2} - 18 x + 5 : \frac{df ( x )}{d x} = 4 x - 18

Compute the slope of

f (x) = 2 x^{2} - 18 x + 5

x = 0 : m = - 18

Find the line with slope m=

- 18

and passing through

(0, 5) : y = - 18 x + 5

y = - 18 x + 5

x \to \infty lim (\frac{3 x ^{2} - x - 2}{5 x ^{2} - 4 x + 1})

f (x) = \frac{x ^{2}}{3 + 8 x}

(\frac{x ^{2}}{2})^{^{'}}

\int_{0}^{4} x 17 - x^{2} d x

\int_{0}^{1} \frac{x ^{4} + 6}{x} d x

What is the tangent of f(x)=2x^2-18x+5,\at x=0 ?
The tangent of f(x)=2x^2-18x+5,\at x=0 is y=-18x+5