What is the axis of symmetry and vertex for the graph $y = (1) {(x - 3)}^{2} + (- 1)$ $y=(1)(x-3)^2+(-1)$ ?

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Answer 1 · 2018-06-21T20:41:36

$axis of symmetry = 3$ $"axis of symmetry" = 3$

$vertex = (3, - 1)$ $"vertex" = (3,-1)$

$y = (1) {(x - 3)}^{2} + (- 1)$ $y=(1)(x-3)^2+(-1)$

$y = {(x - 3)}^{2} - 1$ $y=(x-3)^2-1$

This quadratic equation is in vertex form:

$y = a {(x + h)}^{2} + k$ $y=a(x+h)^2+k$

In this form:

$a = direction parabola opens and stretch$ $a = "direction parabola opens and stretch"$

$vertex = (- h, k)$ $"vertex" = (-h,k)$

$axis of symmetry = - h$ $"axis of symmetry" = -h$

$vertex = (3, - 1)$ $"vertex" = (3,-1)$

$axis of symmetry = 3$ $"axis of symmetry" = 3$

finally, since $a = 1$ $a=1$ , it follows $a > 0$ $a>0$ then vertex is a minimum and the parabola opens up.

graph{y=(x-3)^2-1 [-10, 10, -5, 5]}

What is the axis of symmetry and vertex for the graph y=(1)(x−3)2+(−1)y=(1)(x-3)^2+(-1)?