What is the tangent of 5x^2+4x-7 ?

The tangent of 5x^2+4x-7 is y=(10a_{0}+4)x-5a_{0}^2-7

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tangent of 5x^2+4x-7

tangent of

5 x^{2} + 4 x - 7

y = (10 a_{0} + 4) x - 5 a_{0}^{2} - 7

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 5 a_{0}^{2} + 4 a_{0} - 7)

Find the slope of

y = 5 x^{2} + 4 x - 7 : \frac{d y}{d x} = 10 x + 4

Compute the slope of

y = 5 x^{2} + 4 x - 7

x = a_{0} : m = 10 a_{0} + 4

Find the line with slope m=

10 a_{0} + 4

and passing through

(a_{0}, 5 a_{0}^{2} + 4 a_{0} - 7) : y = (10 a_{0} + 4) x - 5 a_{0}^{2} - 7

y = (10 a_{0} + 4) x - 5 a_{0}^{2} - 7

\frac{d}{d x} (2 + 3 cos (x))

\int x^{3} \cdot e^{- x^{4}} d x

\int cot^{2} (1) d x

y^{^{''}} - 6 y^{^{'}} + 8 y = e^{3 t}

\int \frac{x ^{4} + x - 1}{x ^{2} + 1} d x

What is the tangent of 5x^2+4x-7 ?
The tangent of 5x^2+4x-7 is y=(10a_{0}+4)x-5a_{0}^2-7