What is f(x)-4x^2+12x+27=35 ?

The solution to f(x)-4x^2+12x+27=35 is Parabola with (h,k)=(3/2 ,-1),p= 1/16

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f(x)-4x^2+12x+27=35

f (x) - 4 x^{2} + 12 x + 27 = 35

(h, k) = (\frac{3}{2}, - 1), p = \frac{1}{16}

Solution steps

y - 4 x^{2} + 12 x + 27 = 35

Rewrite

y - 4 x^{2} + 12 x + 27 = 35

in the standard form

4 \cdot \frac{1}{16} (y - (- 1)) = (x - \frac{3}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{3}{2}, - 1), p = \frac{1}{16}

y^{2} + 6 y - x + 8 = 0

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

- x^{2} - 9 x + y - 14 < 0

1 f (x) = - 5 x^{2} + 8

(x + 3)^{2} = - y

What is f(x)-4x^2+12x+27=35 ?
The solution to f(x)-4x^2+12x+27=35 is Parabola with (h,k)=(3/2 ,-1),p= 1/16