What is the tangent of f(x)=2x^2-7x+5,\at x=0 ?

The tangent of f(x)=2x^2-7x+5,\at x=0 is y=-7x+5

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tangent of f(x)=2x^2-7x+5,\at x=0

tangent of

f (x) = 2 x^{2} - 7 x + 5, at x = 0

y = - 7 x + 5

Solution steps

Find the tangent point:

(0, 5)

Find the slope of

f (x) = 2 x^{2} - 7 x + 5 : \frac{df ( x )}{d x} = 4 x - 7

Compute the slope of

f (x) = 2 x^{2} - 7 x + 5

x = 0 : m = - 7

Find the line with slope m=

- 7

and passing through

(0, 5) : y = - 7 x + 5

y = - 7 x + 5

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

\frac{\partial}{\partial x} (\frac{9 x}{x ^{2} + 5 y})

2 x - \frac{1}{2 x}

\int_{0}^{1} 9 x^{2} + 1 d x

\int sec (\frac{2 x}{5}) tan (\frac{2 x}{5}) d x

What is the tangent of f(x)=2x^2-7x+5,\at x=0 ?
The tangent of f(x)=2x^2-7x+5,\at x=0 is y=-7x+5