What is k(x)-1(x)=x^2+5 ?

The solution to k(x)-1(x)=x^2+5 is Parabola with (h,k)=(-1/2 , 19/4),p= 1/4

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k(x)-1(x)=x^2+5

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

(h, k) = (- \frac{1}{2}, \frac{19}{4}), p = \frac{1}{4}

Solution steps

y - 1 \cdot x = x^{2} + 5

Rewrite

y - 1 \cdot x = x^{2} + 5

in the standard form

4 \cdot \frac{1}{4} (y - \frac{19}{4}) = (x - (- \frac{1}{2}))^{2}

Therefore parabola properties are:

(h, k) = (- \frac{1}{2}, \frac{19}{4}), p = \frac{1}{4}

x^{2} - 6 x + 4 y - 11 = 0

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

x^{2} + 14 x - y + 42 = 0

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

x^{2} - 8 y - x + 15 = 0

What is k(x)-1(x)=x^2+5 ?
The solution to k(x)-1(x)=x^2+5 is Parabola with (h,k)=(-1/2 , 19/4),p= 1/4