What is the vertices (y+3)^2=4(x-3) ?

The vertices (y+3)^2=4(x-3) is (h,k)=(3,-3),p=1

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vertices (y+3)^2=4(x-3)

vertices

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

(h, k) = (3, - 3), p = 1

Solution steps

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

Rewrite

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

in the standard form:

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

Therefore parabola properties are:

(h, k) = (3, - 3), p = 1

y = 7 x^{2} - 10 x + 5

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

(x + 2)^{2} - 9

y = x^{2} - 8 x + 7

f (x) = - x^{2} + 7

What is the vertices (y+3)^2=4(x-3) ?
The vertices (y+3)^2=4(x-3) is (h,k)=(3,-3),p=1