Download Completing the square

Completing the square
Consider the parabola y = x 2 4x +1 .
4
y
3
Complete the square on the quadratic:
2
y = x 2 4x +1
1
= (x 4x) +1
2
2
1
2
()
–1
2
1
0
2
3
4
5 x
–1
–2
= (x 4x + 4) +1 4
–3
= (x 2) 3
–4
2
2
(Notice how four was added inside the parentheses to complete the
square, and subtracted outside to maintain the equality.)
Now compare y = (x 2)2 3 with the vertex of the parabola shown in
the graph. Can you see a relationship between the two?
For a parabola written
in the form
y = (x - h)2 + k
the vertex has
coordinates (h, k ).
h
k
vertex (h, k )
The parabola y = (x 2)2 3 has vertex (2, –3). Immediately, you also
know that the axis of symmetry of the parabola is x = 2 .
Why is this so?
Part 1
Parabolas
1
You know that the parabola y = (x h)2 is
the same as the parabola y = x 2 shifted h
5
units horizontally.
3
4
2
So y = (x 2)2 is shifted two units to the
1
right, having vertex (2, 0).
And y = (x h)2 + k is the parabola
y = (x h)2 shifted q units vertically.
So y = (x 2)2 3 is y = (x 2)2 shifted
y
–1
0
1
2
3
4
5 x
–1
–2
–3
–4
three units downwards. It has vertex (2, –
3).
By writing a parabola in the form y = (x h)2 + k you can quickly
determine the coordinates of the vertex.
This is just tying together information you learned previously.
Follow through the steps in this example. Do your own working in the
margin if you wish.
Write y = 4 2x x 2 in the form y = (x h)2 + k and hence
determine the vertex of this parabola. Draw a sketch of this
curve showing the main features.
Solution
y = 4 2x x 2
y = [x 2 + 2x 4]
= [(x 2 + 2x) 4]
= [(x 2 + 2x +1) 4 1] (completing the square)
= [(x +1)2 5]
= (x +1)2 + 5
The vertex is (–1, 5).
(The minus in front of the squared term indicates the parabola
is facing concave down.)
2
PAS5.3.4 Coordinate geometry
From the original y = 4 2x x 2 the y-intercept is 4.
You now have enough information to draw a sketch.
y
5
4
3
2
1
–4
–3
–2
–1
0
1
2 x
–1
–2
In this question you weren’t asked to calculate the x-intercepts. However
from the sketch you can see they lie somewhere between –4 and –3, and
between 1 and 2.
If you needed their values, you could start with y = (x +1)2 + 5 .
(x +1)2 + 5 = 0
(x +1)2 = 5
x +1 = ± 5
x = 1± 5 (exact value)
x = 3.24or1.24(correct to 2 dec. pl.)
Part 1
Parabolas
3
Activity – Completing the square
Try these.
1
Write the vertex to y = (x + 7)2 + 5 . _________________________
2
The parabola y = (x h)2 + k has vertex (–2, 1). What are the
values of h and k? _______________________________________
Write y = x 2 6x 3 in the form y = (x h)2 + k and hence
3
determine the vertex of this parabola.
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
_______________________________________________________
Draw a sketch of this curve showing the main features.
3
y
2
1
–1
0
–1
1
2
3
4
5
6
7
8x
–2
–3
–4
–5
–6
–7
–8
–9
–10
–11
–12
–13
–14
4
PAS5.3.4 Coordinate geometry