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

Jun 21, 2018

$\text{axis of symmetry} = 3$

$\text{vertex} = \left(3 , - 1\right)$

Explanation:

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

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

This quadratic equation is in vertex form:

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

In this form:

$a = \text{direction parabola opens and stretch}$

$\text{vertex} = \left(- h , k\right)$

$\text{axis of symmetry} = - h$

$\text{vertex} = \left(3 , - 1\right)$

$\text{axis of symmetry} = 3$

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

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