What is the tangent of y= 1/3 x^2-2x+4,(3,1) ?

The tangent of y= 1/3 x^2-2x+4,(3,1) is y=1

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tangent of y= 1/3 x^2-2x+4,(3,1)

tangent of

y = \frac{1}{3} x^{2} - 2 x + 4, (3, 1)

y = 1

Solution steps

Find the slope of

y = \frac{1}{3} x^{2} - 2 x + 4 : \frac{d y}{d x} = \frac{2 x}{3} - 2

Compute the slope of

y = \frac{1}{3} x^{2} - 2 x + 4

(3, 1) : m = 0

Find the line with slope m=

0

and passing through

(3, 1) : y = 1

y = 1

\frac{d}{d x} (cos (tan (4 x)))

\frac{d}{d x} (\frac{x ^{3} - 6 x ^{2} + 5}{x})

n = 0 \sum \infty (\frac{1}{6})^{n}

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

\frac{\partial}{\partial y} (3 cos (y))

What is the tangent of y= 1/3 x^2-2x+4,(3,1) ?
The tangent of y= 1/3 x^2-2x+4,(3,1) is y=1