What is the tangent of 4x^2-5x ?

The tangent of 4x^2-5x is y=(8a_{0}-5)x-4a_{0}^2

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tangent of 4x^2-5x

tangent of

4 x^{2} - 5 x

y = (8 a_{0} - 5) x - 4 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} - 5 a_{0})

Find the slope of

y = 4 x^{2} - 5 x : \frac{d y}{d x} = 8 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 = 8 a_{0} - 5

Find the line with slope m=

8 a_{0} - 5

and passing through

(a_{0}, 4 a_{0}^{2} - 5 a_{0}) : y = (8 a_{0} - 5) x - 4 a_{0}^{2}

y = (8 a_{0} - 5) x - 4 a_{0}^{2}

\int \frac{e ^{2 y}}{e ^{2 y} - 1} d y

y^{4} - 4 y^{2} = x^{4} - 9 x^{2}, (3, 2)

\int (3 x^{3} - \frac{1}{4 x}) d x

\int e^{x - 2} d x

n = 0 \sum \infty 9 (\frac{1 - 3}{4})^{n}

What is the tangent of 4x^2-5x ?
The tangent of 4x^2-5x is y=(8a_{0}-5)x-4a_{0}^2