What is the tangent of f(x)=2x^2-5x+4,\at x=0 ?

The tangent of f(x)=2x^2-5x+4,\at x=0 is y=-5x+4

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tangent of f(x)=2x^2-5x+4,\at x=0

tangent of

f (x) = 2 x^{2} - 5 x + 4, at x = 0

y = - 5 x + 4

Solution steps

Find the tangent point:

(0, 4)

Find the slope of

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

Compute the slope of

f (x) = 2 x^{2} - 5 x + 4

x = 0 : m = - 5

Find the line with slope m=

- 5

and passing through

(0, 4) : y = - 5 x + 4

y = - 5 x + 4

n = 1 \sum \infty \frac{3 n}{4 n + 5}

\int \frac{1}{- sec ^{2} ( x )} d x

(4 x^{3} + 4 x y - 1) d x = (1 - 2 x^{2} - 2 y) d y

\frac{d}{d x} (∣ x ∣^{∣ x ∣})

x \frac{d y}{d x} + y = - 9 x y^{2}

What is the tangent of f(x)=2x^2-5x+4,\at x=0 ?
The tangent of f(x)=2x^2-5x+4,\at x=0 is y=-5x+4