What is the tangent of f(x)=9-4x+6x^2,\at x=8 ?

The tangent of f(x)=9-4x+6x^2,\at x=8 is y=92x-375

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tangent of f(x)=9-4x+6x^2,\at x=8

tangent of

f (x) = 9 - 4 x + 6 x^{2}, at x = 8

y = 92 x - 375

Solution steps

Find the tangent point:

(8, 361)

Find the slope of

f (x) = 9 - 4 x + 6 x^{2} : \frac{df ( x )}{d x} = 12 x - 4

Compute the slope of

f (x) = 9 - 4 x + 6 x^{2}

x = 8 : m = 92

Find the line with slope m=

92

and passing through

(8, 361) : y = 92 x - 375

y = 92 x - 375

\frac{\partial}{\partial y} (y^{2} e^{x y^{2}})

\int z sin (x) d y

\frac{1}{3} x^{3} + \frac{7}{2} x^{2} - 3 x + 5

f (t) = tan (t)

x \to 1 lim (e^{x^{2} - x})

What is the tangent of f(x)=9-4x+6x^2,\at x=8 ?
The tangent of f(x)=9-4x+6x^2,\at x=8 is y=92x-375