What is the tangent of f(x)=2x^2-5x+3 ?

The tangent of f(x)=2x^2-5x+3 is y=(4a_{0}-5)x-2a_{0}^2+3

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tangent of f(x)=2x^2-5x+3

tangent of

f (x) = 2 x^{2} - 5 x + 3

y = (4 a_{0} - 5) x - 2 a_{0}^{2} + 3

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} + 3)

Find the slope of

f (x) = 2 x^{2} - 5 x + 3 : \frac{df ( x )}{d x} = 4 x - 5

Compute the slope of

f (x) = 2 x^{2} - 5 x + 3

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} + 3) : y = (4 a_{0} - 5) x - 2 a_{0}^{2} + 3

y = (4 a_{0} - 5) x - 2 a_{0}^{2} + 3

x \to 0 + lim (ln (1 + x))

y = 5 tan (x), (\frac{π}{4}, 5)

f (x) = \frac{2}{x}, (- 4, - \frac{1}{2})

\frac{d}{d x} (x^{2} (x + 3)^{4})

n = 1 \sum \infty \frac{5 ^{n}}{6 ^{n} n}

What is the tangent of f(x)=2x^2-5x+3 ?
The tangent of f(x)=2x^2-5x+3 is y=(4a_{0}-5)x-2a_{0}^2+3