What is the tangent of f(x)=2x^2-5x+3,\at x=0 ?

The tangent of f(x)=2x^2-5x+3,\at x=0 is y=-5x+3

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tangent of f(x)=2x^2-5x+3,\at x=0

tangent of

f (x) = 2 x^{2} - 5 x + 3, at x = 0

y = - 5 x + 3

Solution steps

Find the tangent point:

(0, 3)

Find the slope of

f (x) = 2 x^{2} - 5 x + 3 : \frac{df ( x )}{d x} = 4 x - 5

Compute the slope of

f (x) = 2 x^{2} - 5 x + 3

x = 0 : m = - 5

Find the line with slope m=

- 5

and passing through

(0, 3) : y = - 5 x + 3

y = - 5 x + 3

\int \frac{8 + x + x}{x} d x

\frac{d x}{d t} = x sin (t)

x \to 1 - lim (arcsin (x))

\int x^{2} x^{3} + 4 d x

y^{^{'}} - 4 y = 6 e^{6 t}

What is the tangent of f(x)=2x^2-5x+3,\at x=0 ?
The tangent of f(x)=2x^2-5x+3,\at x=0 is y=-5x+3