Comparison to the graph of f(x)=x^2: There is a vertical shrink by a factor of 12, followed by a translation right 2 units and down 4 units of the graph of f.
Practice makes perfect
To graph the given parabola, that is written in the vertex form, we first need to identify the axis of symmetry and the vertex.
f(x)= a (x- h)^2 + k
In this form, the axis of symmetry is the vertical line x= h and the vertex lies at ( h, k). Now consider the given function.
r(x)=1/2(x-2)^2-4
⇕
r(x)= 1/2(x- 2)^2 + ( - 4)
We can see that a= 1/2, h= 2 and k= - 4. Therefore, the vertex is ( 2, - 4), and the axis of symmetry is x= 2. To graph the function we need to find two more points on the graph. Let's choose two x-values less than the x-coordinate of the vertex and make a table of values.
x
1/2(x-2)^2-4
r(x)=1/2(x-2)^2-4
0
1/2 ( 0-2)^2-4
- 2
- 2
1/2 ( - 2-2)^2-4
4
Next, we will plot the vertex and draw the axis of symmetry on a coordinate plane. We will also plot and reflect the obtained points across the axis of symmetry.
Let's draw a smooth curve that connects the five obtained points. We will also draw the parent function f(x)=x^2.
From the graph above, we can note the following.
Both graphs open up.
The graph of r(x)= 12(x-2)^2-4 is wider than the graph of f(x)=x^2.
The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=2.
The vertex of the given function, (2,- 4), is below and to the right of the vertex of the parent function, (0,0).
From the graph and the observations above, we can conclude that the graph of r is a vertical shrink by a factor of 12, followed by a translation right 2 units and down 4 units of the graph of f.
