What is the tangent of f(x)=(x^2)/(x+2),\at x=2 ?

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

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tangent of f(x)=(x^2)/(x+2),\at x=2

tangent of

f (x) = \frac{x ^{2}}{x + 2}, at x = 2

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

Solution steps

Find the tangent point:

(2, 1)

Find the slope of

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

Compute the slope of

f (x) = \frac{x ^{2}}{x + 2}

x = 2 : m = \frac{3}{4}

Find the line with slope m=

\frac{3}{4}

and passing through

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

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

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

\frac{d}{d x} (4 - x^{2} + 2 arcsin (\frac{x}{2}))

\frac{x ^{2}}{x - 3}

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

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