What is the vertices 4x-y^2+y-31/4 =0 ?

The vertices 4x-y^2+y-31/4 =0 is (h,k)=(15/8 , 1/2),p=1

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vertices 4x-y^2+y-31/4 =0

vertices

4 x - y^{2} + y - \frac{31}{4} = 0

(h, k) = (\frac{15}{8}, \frac{1}{2}), p = 1

Solution steps

4 x - y^{2} + y - \frac{31}{4} = 0

Rewrite

4 x - y^{2} + y - \frac{31}{4} = 0

in the standard form

4 \cdot 1 \cdot (x - \frac{15}{8}) = (y - \frac{1}{2})^{2}

Therefore parabola properties are:

(h, k) = (\frac{15}{8}, \frac{1}{2}), p = 1

f (x) = x^{2} - 10 x + 16

- 3 x^{2} - 5

f (x) = x^{2} - 8 x - 9

X^{2} + 2 X - 8

f (x) = x^{2} - 4 x - 4

What is the vertices 4x-y^2+y-31/4 =0 ?
The vertices 4x-y^2+y-31/4 =0 is (h,k)=(15/8 , 1/2),p=1