Question

1. y=(x - 2)^2

vertex:
axis of symmetry: x=
opens: up down
maximum minimum
max/min value:
y - intercept:

Explanation:

Step1: Identify the vertex - form of parabola

The equation of the parabola is $y=(x - 2)^2$, which is in the vertex - form $y=a(x - h)^2+k$, where $(h,k)$ is the vertex. Here $h = 2$ and $k = 0$, so the vertex is $(2,0)$.

Step2: Find the axis of symmetry

For a parabola in the form $y=a(x - h)^2+k$, the axis of symmetry is given by the equation $x=h$. Since $h = 2$, the axis of symmetry is $x = 2$.

Step3: Determine the opening direction

In the equation $y=(x - 2)^2$, $a = 1>0$, so the parabola opens up. When a parabola opens up, it has a minimum value.

Step4: Find the minimum value

The minimum value of a parabola $y=a(x - h)^2+k$ occurs at the vertex. Since the vertex is $(2,0)$, the minimum value of the function is $y = 0$.

Step5: Find the y - intercept

To find the y - intercept, set $x = 0$ in the equation $y=(x - 2)^2$. Then $y=(0 - 2)^2=4$.

Answer:

Vertex: $(2,0)$
Axis of symmetry: $x = 2$
Opens: up
Maximum/Minimum: Minimum
Min value: $y = 0$
y - intercept: $(0,4)$