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

The tangent of f(x)=3x^2-4x+6,\at x=2 is y=8x-6

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

tangent of

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

y = 8 x - 6

Solution steps

Find the tangent point:

(2, 10)

Find the slope of

f (x) = 3 x^{2} - 4 x + 6 : \frac{df ( x )}{d x} = 6 x - 4

Compute the slope of

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

x = 2 : m = 8

Find the line with slope m=

8

and passing through

(2, 10) : y = 8 x - 6

y = 8 x - 6

f (x) = x^{3} - 3 x + 4, (3, 22)

\frac{d}{d x} (1 + x^{4})

f (x) = \frac{x ^{2}}{x - 8}

\int 2 x^{2} y e^{x yz} d x

\frac{\partial}{\partial x} (- 2 x z)

What is the tangent of f(x)=3x^2-4x+6,\at x=2 ?
The tangent of f(x)=3x^2-4x+6,\at x=2 is y=8x-6