Question

Identify the vertex and the axis of symmetry of the graph of the function y=3(x+2)^2-3

Answer 1

The vertex of the function $y=3{\left({x+2}\right)^2}-3$ is $\boxed{\left({-2,-3}\right)}$.

Further Explanation:

The standard form of the parabola is shown below.

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

Here, the parabola has vertex at $\left({h,k}\right)$ and has the symmetry parallel to x-axis and it opens left.

Given:

The quadratic function is $y=3{\left({x+2}\right)^2}-3$.

Calculation:

Compare the $y=3{\left({x+2}\right)^2}-3$ with the general equation of the parabola $\boxed{y=a{{\left({x-h}\right)}^2}+k}$

.

The value $a$ is $3$, the value of $h$ is $-2$ and the value of $k$ is $-3$.

Therefore, the vertex of the parabola is $\left({-2,-3}\right)$.

The function is symmetric about $x=-2$.

The vertex of the function $y=3{\left({x+2}\right)^2}-3$ is $\boxed{\left({-2,-3}\right)}$.

Learn more:

1. Learn more about unit conversion brainly.com/question/4837736

2. Learn more about non-collinear brainly.com/question/4165000

3. Learn more about binomial and trinomial brainly.com/question/1394854

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Conic sections

Keywords: vertex, symmetry, symmetric, axis, y-axis, x-axis, function, graph, parabola, focus, vertical parabola, upward parabola, downward parabola,

Answer 2

we know that

The equation in vertex form of a vertical parabola is of the form

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

where

$(h,k)$ is the vertex of the parabola

and

$x=h$ is the axis of symmetry

if $a > 0$ -----> open upward

if $a < 0$ -----> open downward

In this problem we have

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

$a=3$

This is a vertical parabola open upward

The vertex is a minimum

therefore

the answer is

the vertex is the point $(-2,-3)$

the axis of symmetry is $x=-2$

see the attached figure to better understand the problem