Range of f(x) = -(x + 2)^2 + 1 Step by Step

Step by step

The question

Question: find the range of f(x) = −(x + 2)² + 1.

f(x) = −(x + 2)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 + 2)² is a square too.

So it is 0 or more.

(x + 2)² ≥ 0

−(x + 2)² is minus a square.

So it is 0 or less.

It is never above 0.

−(x + 2)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.

−(x + 2)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 down.

y ≤ 1

3 See it on the graph

On the graph, the turning point is (−2, 1).

The curve never goes above y = 1.

Its highest y is 1.

4 Check

Let's check.

Put in x = −2, the turning point.

We get 1, the biggest y.

x = −2: y = −((−2) + 2)² + 1 = −(0)² + 1

= 0 + 1 = 1

Now x = 0.

We get −3.

−3 is less than 1.

It fits.

x = 0: y = −(0 + 2)² + 1 = −(2)² + 1

= −4 + 1 = −3

Let's look back.

First: what a square can be.

Then: add the 1.

Then: see it on the graph.

Last: check.

Answer