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

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

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

tangent of

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

y = 2 x^{2} - 5 x + 3 : \frac{d y}{d x} = 4 x - 5

Compute the slope of

y = 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

x^{t a n (x)}

\frac{d y}{d x} = cos (y - x)

\int_{0}^{N} e^{- s t} sin (5 t) d t

\frac{1}{s ( 2 s + 1 )}

\frac{d}{d x} (se x + cos (x))

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