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

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

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tangent of 2x^2-5x

tangent of

2 x^{2} - 5 x

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Find the line with slope m=

4 a_{0} - 5

and passing through

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

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

f (x) = (2 x - 9)^{4} (x^{2} + x + 1)^{5}

y = x^{2} - 2 x, (5, 15)

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

\int \frac{sin ( x )}{cos ( x ) + 1} d x

(y - y x^{2}) \frac{d y}{d x} = (y + 1)^{2}

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