y = (x - 4)^2 + 1: which form shows the lowest point Step by Step

Step by step

The question

Question: which form of y = (x − 4)² + 1 shows the lowest point?

y = (x − 4)² + 1

1 Pick the form

A parabola can be written in three forms.

Each one shows one thing at a glance.

The vertex form has a square, and a square is never below 0.

So it shows the lowest point.

standard: ax² + bx + c → y axis at c

factored: a(x − r)(x − s) → x axis at r and s

vertex: a(x − h)² + k → turning point (h, k) ✓

It is already in vertex form.

2 Read the lowest point

A square is never below 0: a number times itself is never minus.

(x − 4) squared is smallest, 0, when x − 4 is 0: at x = 4.

(x − 4)² is never below 0; it is 0 when x = 4

There the square part is 0.

So y is just 1.

Anywhere else we add a bit more than 0, so y is bigger.

So this is the lowest point.

x = 4: y = 0 + 1 = 1

Careful!

Not −4.

x − 4 is 0 when x is 4: the sign in the bracket is the opposite.

Careful: not (x − 4)²: x = −4 ✗

So the lowest point is (4, 1).

lowest = (4, 1)

3 Show it on the graph

Here is the graph of y = x² − 8x + 17.

Each square up is 2.

The lowest point is (4, 1).

4 Check

Let's check in the question's own form, y = (x − 4)² + 1.

x = 4 gives y = 1, the y of the lowest point.

So it is right!

x = 4: y = (4 − 4)² + 1 = 0² + 1 = 1

Let's look back.

First: pick the form.

Then: read the lowest point.

Then: show it on the graph.

Last: check.

Answer