Question

What is the axis of symmetry and vertex for the function f(x) = 3(x – 2)2 4?

Answer 1

we know that

The equation of a vertical parabola in vertex form is equal to

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

where

(h,k) is the vertex

if $a > 0$ -----> then the parabola open upward (the vertex is a minimum)

if $a < 0$ -----> then the parabola open downward (the vertex is a maximum)

The axis of symmetry is equal to

$x=h$

In this problem let's analyze two cases

First case

$f(x)=3(x-2)^{2}+4$

the vertex is the point $(2,4)$

$a=3$

so

$3 > 0$ -----> then the parabola open upward (the vertex is a minimum)

The axis of symmetry is equal to

$x=2$

Second case

$f(x)=3(x-2)^{2}-4$

the vertex is the point $(2,-4)$

$a=3$

so

$3 > 0$ -----> then the parabola open upward (the vertex is a minimum)

The axis of symmetry is equal to

$x=2$

Answer 2

Hello,

y=3(x-2)²+4

Vertex is(2,4)
Axis of symmetry: x=2