What is the tangent of 7x^2+8x ?

The tangent of 7x^2+8x is y=(14a_{0}+8)x-7a_{0}^2

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tangent of 7x^2+8x

tangent of

7 x^{2} + 8 x

y = (14 a_{0} + 8) x - 7 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 7 a_{0}^{2} + 8 a_{0})

Find the slope of

y = 7 x^{2} + 8 x : \frac{d y}{d x} = 14 x + 8

Compute the slope of

y = 7 x^{2} + 8 x

x = a_{0} : m = 14 a_{0} + 8

Find the line with slope m=

14 a_{0} + 8

and passing through

(a_{0}, 7 a_{0}^{2} + 8 a_{0}) : y = (14 a_{0} + 8) x - 7 a_{0}^{2}

y = (14 a_{0} + 8) x - 7 a_{0}^{2}

\int \frac{x ^{2}}{x ^{2} + 4 x - 12} d x

\int (8 x^{3} + \frac{1}{2 x ^{2}}) d x

\frac{( 3 s - 4 )}{s ^{2} + 9}

\int_{0}^{9} 2 y d y

\frac{d}{d θ} (sec (θ) tan (θ))

What is the tangent of 7x^2+8x ?
The tangent of 7x^2+8x is y=(14a_{0}+8)x-7a_{0}^2