What is the tangent of f(x)=7x^2+5x-8 ?

The tangent of f(x)=7x^2+5x-8 is y=(14a_{0}+5)x-7a_{0}^2-8

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tangent of f(x)=7x^2+5x-8

tangent of

f (x) = 7 x^{2} + 5 x - 8

y = (14 a_{0} + 5) x - 7 a_{0}^{2} - 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 7 a_{0}^{2} + 5 a_{0} - 8)

Find the slope of

f (x) = 7 x^{2} + 5 x - 8 : \frac{df ( x )}{d x} = 14 x + 5

Compute the slope of

f (x) = 7 x^{2} + 5 x - 8

x = a_{0} : m = 14 a_{0} + 5

Find the line with slope m=

14 a_{0} + 5

and passing through

(a_{0}, 7 a_{0}^{2} + 5 a_{0} - 8) : y = (14 a_{0} + 5) x - 7 a_{0}^{2} - 8

y = (14 a_{0} + 5) x - 7 a_{0}^{2} - 8

x \to 0 lim (\frac{3}{sin ( 3 x )})

e^{- x}

y^{^{'}} = 2 x + 4 y

\int \frac{x}{2 + 3 x} d x

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

What is the tangent of f(x)=7x^2+5x-8 ?
The tangent of f(x)=7x^2+5x-8 is y=(14a_{0}+5)x-7a_{0}^2-8