What is the vertex of the parabola $y=3(x-4)^2-22$ ?

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Answer 1 · 2016-05-10T23:20:38

$(4, -22)$

The equation:

$y = 3(x-4)^2-22$

is in vertex form:

$y = a(x-h)+k$

with multiplier $a = 3$ and vertex $(h, k) = (4, -22)$

The nice thing about vertex form is that you can immediately read the vertex coordinates from it.

Notice that $(x-4)^2 >= 0$ , taking its minimum value $0$ when $x=4$ . When $x=4$ we have $y = 3(4-4)^2-22 = 0-22 = -22$ .

So the vertex is at $(4, -22)$ .

What is the vertex of the parabola y=3(x-4)^2-22y=3(x-4)^2-22?