What is f(x)(x)^{-1}=2x+1 ?

The solution to f(x)(x)^{-1}=2x+1 is Parabola with (h,k)=(-1/4 ,-1/8),p= 1/8

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f(x)(x)^{-1}=2x+1

f (x) (x)^{- 1} = 2 x + 1

(h, k) = (- \frac{1}{4}, - \frac{1}{8}), p = \frac{1}{8}

Solution steps

y x^{- 1} = 2 x + 1

Rewrite

y x^{- 1} - 2 x - 1 = 0

in the standard form

4 \cdot \frac{1}{8} (y - (- \frac{1}{8})) = (x - (- \frac{1}{4}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{4}, - \frac{1}{8}), p = \frac{1}{8}

y^{2} = 3 x - 5

3 y^{2} - 8 x + 5 y - 13 = 0

y^{(2)} = \frac{( 18 - 3 x )}{( 2 )}

\frac{1}{3} π r^{2} h = v

f (x) - 5 = x^{2} - 3 x + 2

What is f(x)(x)^{-1}=2x+1 ?
The solution to f(x)(x)^{-1}=2x+1 is Parabola with (h,k)=(-1/4 ,-1/8),p= 1/8