What is the tangent of f(x)=2x^2+x-1 ?

The tangent of f(x)=2x^2+x-1 is y=(4a_{0}+1)x-2a_{0}^2-1

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tangent of f(x)=2x^2+x-1

tangent of

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

y = (4 a_{0} + 1) x - 2 a_{0}^{2} - 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2} + a_{0} - 1)

Find the slope of

f (x) = 2 x^{2} + x - 1 : \frac{df ( x )}{d x} = 4 x + 1

Compute the slope of

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

x = a_{0} : m = 4 a_{0} + 1

Find the line with slope m=

4 a_{0} + 1

and passing through

(a_{0}, 2 a_{0}^{2} + a_{0} - 1) : y = (4 a_{0} + 1) x - 2 a_{0}^{2} - 1

y = (4 a_{0} + 1) x - 2 a_{0}^{2} - 1

y^{^{''}} + 9 y = 0, y (\frac{π}{2}) = 1, y^{^{'}} (\frac{π}{2}) = - 3

\frac{d y}{d x}, 9 y + x = 4 y

\int (6 x^{2} - 4 e^{x}) d x

\frac{5 x ^{3} + 5 x ^{2}}{x}

\int \frac{arctan ( x )}{x ^{2}} d x

What is the tangent of f(x)=2x^2+x-1 ?
The tangent of f(x)=2x^2+x-1 is y=(4a_{0}+1)x-2a_{0}^2-1