What is the tangent of 4x^2-3x-5 ?

The tangent of 4x^2-3x-5 is y=(8a_{0}-3)x-4a_{0}^2-5

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tangent of 4x^2-3x-5

tangent of

4 x^{2} - 3 x - 5

y = (8 a_{0} - 3) 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} - 3 a_{0} - 5)

Find the slope of

y = 4 x^{2} - 3 x - 5 : \frac{d y}{d x} = 8 x - 3

Compute the slope of

y = 4 x^{2} - 3 x - 5

x = a_{0} : m = 8 a_{0} - 3

Find the line with slope m=

8 a_{0} - 3

and passing through

(a_{0}, 4 a_{0}^{2} - 3 a_{0} - 5) : y = (8 a_{0} - 3) x - 4 a_{0}^{2} - 5

y = (8 a_{0} - 3) x - 4 a_{0}^{2} - 5

\int \frac{( x ^{2} + 1 ) ^{2}}{x ^{2}} d x

\int cos (x) (e^{7 s i n (x)}) d x

\frac{( - 8 x )}{2 x - 1}

y^{^{'}} = \frac{2 x y ^{2} + 8 x}{( x ^{2} y + 3 y )}

x \to 0 + lim ((e^{x} - 1)^{x})

What is the tangent of 4x^2-3x-5 ?
The tangent of 4x^2-3x-5 is y=(8a_{0}-3)x-4a_{0}^2-5