What is the vertices x<= y^2 ?

The vertices x<= y^2 is (h,k)=(0,0),p= 1/4

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vertices x<= y^2

vertices

x \leq y^{2}

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

Solution steps

x \leq y^{2}

To find the parabola properties rewrite as parabola:

x = y^{2}

Rewrite

x = y^{2}

in the standard form:

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

Therefore parabola properties are:

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

\frac{1}{2} x^{2} + 8 x + 5

y = 2 x^{2} - 12 x + 3

f (x) = x^{2} - 8 x + 15

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

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