What is the tangent of f(x)=6x^2+4x-7,-3 ?

The tangent of f(x)=6x^2+4x-7,-3 is y=(12a_{0}+4)x-6a_{0}^2-7

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tangent of f(x)=6x^2+4x-7,-3

tangent of

f (x) = 6 x^{2} + 4 x - 7, - 3

y = (12 a_{0} + 4) x - 6 a_{0}^{2} - 7

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 6 a_{0}^{2} + 4 a_{0} - 7)

Find the slope of

f (x) = 6 x^{2} + 4 x - 7 : \frac{df ( x )}{d x} = 12 x + 4

Compute the slope of

f (x) = 6 x^{2} + 4 x - 7

x = a_{0} : m = 12 a_{0} + 4

Find the line with slope m=

12 a_{0} + 4

and passing through

(a_{0}, 6 a_{0}^{2} + 4 a_{0} - 7) : y = (12 a_{0} + 4) x - 6 a_{0}^{2} - 7

y = (12 a_{0} + 4) x - 6 a_{0}^{2} - 7

\int \frac{( arcsin ( \frac{x}{3} ) ) ^{2}}{9 - x ^{2}} d x

\frac{\partial}{\partial y} ((x - y)^{5})

\frac{d}{d x} (\frac{5 ^{x}}{ln ( 5 )})

\int (e - x - e^{2}) (e x + e - 5) d x

\int 3 e^{3 x} d x

What is the tangent of f(x)=6x^2+4x-7,-3 ?
The tangent of f(x)=6x^2+4x-7,-3 is y=(12a_{0}+4)x-6a_{0}^2-7