What is the tangent of f(x)=4x^2-2x+4,\at x=0 ?

The tangent of f(x)=4x^2-2x+4,\at x=0 is y=-2x+4

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tangent of f(x)=4x^2-2x+4,\at x=0

tangent of

f (x) = 4 x^{2} - 2 x + 4, at x = 0

y = - 2 x + 4

Solution steps

Find the tangent point:

(0, 4)

Find the slope of

f (x) = 4 x^{2} - 2 x + 4 : \frac{df ( x )}{d x} = 8 x - 2

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 = - 2

Find the line with slope m=

- 2

and passing through

(0, 4) : y = - 2 x + 4

y = - 2 x + 4

5 - 9 + 8 x^{2}

\int sec^{2} (2 x) tan^{4} (2 x) d x

\int \frac{1}{5 x ^{2} + 1} d x

x \to + 100 lim (7)

\int \frac{4 + x - 3 x}{x} d x

What is the tangent of f(x)=4x^2-2x+4,\at x=0 ?
The tangent of f(x)=4x^2-2x+4,\at x=0 is y=-2x+4