What is the vertices (x-2)=3(y-5)^2 ?

The vertices (x-2)=3(y-5)^2 is (h,k)=(2,5),p= 1/12

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vertices (x-2)=3(y-5)^2

vertices

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

(h, k) = (2, 5), p = \frac{1}{12}

Solution steps

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

Rewrite

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

in the standard form:

4 \cdot \frac{1}{12} (x - 2) = (y - 5)^{2}

Therefore parabola properties are:

(h, k) = (2, 5), p = \frac{1}{12}

y = \frac{x ^{2} x ^{1}}{x}

X^{2}

3 - x^{2}

y = x^{2} - 5 x + 6

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

What is the vertices (x-2)=3(y-5)^2 ?
The vertices (x-2)=3(y-5)^2 is (h,k)=(2,5),p= 1/12