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

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

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tangent of f(x)=4x^2-2x+5

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = 4 x^{2} - 2 x + 5 : \frac{df ( x )}{d x} = 8 x - 2

Compute the slope of

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

x = a_{0} : m = 8 a_{0} - 2

Find the line with slope m=

8 a_{0} - 2

and passing through

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

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

(sec^{2} (x))^{^{'}}

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

x \to 0 + lim (tan^{x} (9 x))

\frac{d}{d x} (π^{c o s (x)})

x \to 0 lim ((\frac{x + 2}{x})^{2} - 4)

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