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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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

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

x \to \frac{π}{3} lim (sin (3 x))

x \to \infty lim (x^{2} - 3)

d er i v a t i v e 7 x^{4}

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