What is the tangent of f(x)=2x^2+8x-2 ?

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

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tangent of f(x)=2x^2+8x-2

tangent of

f (x) = 2 x^{2} + 8 x - 2

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2} + 8 a_{0} - 2)

Find the slope of

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

Compute the slope of

f (x) = 2 x^{2} + 8 x - 2

x = a_{0} : m = 4 a_{0} + 8

Find the line with slope m=

4 a_{0} + 8

and passing through

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

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

\frac{7}{x ^{2}}

x \to \infty lim (\frac{4 ^{x}}{5 ^{x + 1}})

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

\int (x^{2} - 9 x + 2) d x

\frac{d ^{2}}{d x ^{2}} y + 25 y = 0

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