What is f(x)(b-1)=x^2+7 ?

The solution to f(x)(b-1)=x^2+7 is Parabola with (h,k)=(0, 7/(b-1)),p=(b-1)/4

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f(x)(b-1)=x^2+7

f (x) (b - 1) = x^{2} + 7

(h, k) = (0, \frac{7}{b - 1}), p = \frac{b - 1}{4}

Solution steps

y (b - 1) = x^{2} + 7

Rewrite

y (b - 1) = x^{2} + 7

in the standard form:

4 \cdot \frac{b - 1}{4} (y - \frac{7}{b - 1}) = (x - 0)^{2}

Therefore parabola properties are:

(h, k) = (0, \frac{7}{b - 1}), p = \frac{b - 1}{4}

(x - 4)^{2} = - 4 (- 5) (y + 2)

\frac{1}{t + 5} = \frac{6}{r ^{2} - 25}

y^{2} - 8 y + 8 x + 16 = 0

25 = \frac{x}{y ^{2}}

x^{2} - 76 x + 20 y + 50 = 0

What is f(x)(b-1)=x^2+7 ?
The solution to f(x)(b-1)=x^2+7 is Parabola with (h,k)=(0, 7/(b-1)),p=(b-1)/4