Step by step
The question
Question: find the range of f(x) = 2(x − 3)² + 1.
f(x) = 2(x − 3)2 + 1
The range is all the answers, the y's, that can come out.
range = the y's that can come out
1 What a square can be
A square is never below 0.
3 squared is 9.
Minus 3 squared is 9.
0 squared is 0.
3² = 9, (−3)² = 9, 0² = 0
(x − 3)² is a square too.
So it is 0 or more.
(x − 3)2 ≥ 0
2 times a square is 0 or more too.
2(x − 3)2 ≥ 0
2 Add the 1
In f(x), + 1 comes after.
So we add 1.
We do it on both sides, so the sign stays true.
2(x − 3)2 + 1 ≥ 0 + 1
0 plus 1 is 1.
The left side is f(x), the y.
So y ≥ 1.
The range is every y from 1 up.
y ≥ 1
3 See it on the graph
On the graph, the turning point is (3, 1).
The curve never goes below y = 1.
Its lowest y is 1.
4 Check
Let's check.
Put in x = 3, the turning point.
We get 1, the smallest y.
x = 3: y = 2(3 − 3)2 + 1 = 2(0)2 + 1
= 2 × 0 + 1 = 0 + 1 = 1
Now x = 5.
We get 9.
9 is more than 1.
It fits.
x = 5: y = 2(5 − 3)2 + 1 = 2(2)2 + 1
= 2 × 4 + 1 = 8 + 1 = 9
Let's look back.
First: what a square can be.
Then: add the 1.
Then: see it on the graph.
Last: check.
Answer
y ≥ 1