What is the tangent of 2x^2-5 ?

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

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tangent of 2x^2-5

tangent of

2 x^{2} - 5

y = 4 a_{0} x - 2 a_{0}^{2} - 5

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = 2 x^{2} - 5 : \frac{d y}{d x} = 4 x

Compute the slope of

y = 2 x^{2} - 5

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

Find the line with slope m=

4 a_{0}

and passing through

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

y = 4 a_{0} x - 2 a_{0}^{2} - 5

h (x) = (x - 2) (2 x + 3)

\int 3 - e^{2 x} d x

\frac{d}{d x} (2 ln (2 x^{3}))

x^{2} = e^{(x - y)}

f (t) = \frac{1}{t}