What is the vertices 0=(y-1)^2-16(x+4) ?

The vertices 0=(y-1)^2-16(x+4) is (h,k)=(-4,1),p=4

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vertices 0=(y-1)^2-16(x+4)

vertices

0 = (y - 1)^{2} - 16 (x + 4)

(h, k) = (- 4, 1), p = 4

Solution steps

0 = (y - 1)^{2} - 16 (x + 4)

Rewrite

0 = (y - 1)^{2} - 16 (x + 4)

in the standard form:

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

Therefore parabola properties are:

(h, k) = (- 4, 1), p = 4

y = + 8 - 9 x^{2}

2 π r^{2}

f (x) = - 3 x^{2} + 6 x

y = 14 (x - 4) (x + 10)

x^{2} + 8 x - 80

What is the vertices 0=(y-1)^2-16(x+4) ?
The vertices 0=(y-1)^2-16(x+4) is (h,k)=(-4,1),p=4