What is the tangent of 4x^2-15 ?

The tangent of 4x^2-15 is y=8a_{0}x-4a_{0}^2-15

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tangent of 4x^2-15

tangent of

4 x^{2} - 15

y = 8 a_{0} x - 4 a_{0}^{2} - 15

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0}^{2} - 15)

Find the slope of

y = 4 x^{2} - 15 : \frac{d y}{d x} = 8 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 8 a_{0}

Find the line with slope m=

8 a_{0}

and passing through

(a_{0}, 4 a_{0}^{2} - 15) : y = 8 a_{0} x - 4 a_{0}^{2} - 15

y = 8 a_{0} x - 4 a_{0}^{2} - 15

f (x) = \frac{7}{ln ( x )}

\int \frac{( 4 x + 10 )}{x ^{2} + 2 x + 5} d x

n = 1 \sum \infty \frac{4}{3 ^{n} + 1}

y = 3 x^{2} + 2^{2} x^{2} + 1, (- 1, 0)

\int \frac{- 1}{x + 4} d x

What is the tangent of 4x^2-15 ?
The tangent of 4x^2-15 is y=8a_{0}x-4a_{0}^2-15