What is the tangent of f(x)=2x^2-5x ?

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

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tangent of f(x)=2x^2-5x

tangent of

f (x) = 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

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

Compute the slope of

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

x = a_{0} : 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}

\frac{d}{d x} (- 4 sin (x) + 9 x)

x \to 0 lim (\frac{- 3}{x})

\int_{1}^{\infty} \frac{1}{x ^{1.001}} d x

x \to 4 + lim (\frac{∣ x - 4 ∣}{x - 4})

\frac{d}{d x} ((a x + b)^{5})

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