Download The axis of symmetry

The axis of symmetry
A parabola is symmetrical about its axis of symmetry.
axis of symmetry
axis of symmetry
vertex
vertex
The vertex is the turning point where the axis of symmetry cuts the
parabola.
•
For a parabola that is concave up, the vertex is a minimum turning
point.
•
For a parabola that is concave down, the vertex is a maximum
turning point.
Any horizontal line drawn across the parabola
will intersect the parabola at equal distances on
either side of the axis of symmetry.
So you only really need to draw half of the
parabola on one side of its axis, and then reflect
this across the axis to complete the parabola.
You can locate the axis of symmetry of a parabola by drawing a
horizontal interval across the parabola. The axis runs vertically through
the midpoint of that interval.
Part 1
Parabolas
1
Follow through the steps in this example. Do your own working in the
margin if you wish.
a
What is the equation of the axis of symmetry of the
parabola y = x 2 4x + 8 shown?
(0, 8)
b
(4, 8)
What are the coordinates of its vertex?
Solution
a
The axis runs through the midpoint of (0, 8) and (4, 8).
You can easily see this point to be (2, 8). (You don’t need
to use the midpoint formula for this.) The equation for the
axis of symmetry is x = 2 .
b
The vertex has x-value of 2. Substituting this into the
equation of the parabola,
y = x 2 4x + 8
y = (2)2 4(2) + 8
= 48+8
=4
The vertex is (2, 4).
2
PAS5.3.4 Coordinate geometry
Activity – The axis of symmetry
Try these.
1
a
Show that the points (1, 4) and ( 2 12 , 4) lie on the parabola
y = 2x 2 3x + 9 .
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b
What is the equation of the axis of symmetry?
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c
What are the coordinates of the vertex for this parabola?
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d
Find the x- and y-intercepts for this parabola.
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e
Part 1
Parabolas
Sketch this parabola showing the above features.
3
Activity – The axis of symmetry
1
Substitute both these points into the equation of the parabola,
showing each time the result is true.
a
y = 2x 2 3x + 9
4 = 2(1)2 3(1) + 9
4 = 2 3 + 9
4=4
As this is true, (1, 4) lies on the parabola.
y = 2x 2 3x + 9
4 = 2(2 12 )2 3(2 12 ) + 9
4 = 12 12 + 7 12 + 9
4=4
1
As this is true, (2 ,4) lies on the parabola.
2
b
The axis of symmetry lies at the midpoint of (1, 4) and
1
3
(2 ,4) , which is ( ,4) . The equation of the axis of
2
4
3
symmetry is x = .
4
c
To find the vertex, substitute x = 3
into y = 2x 2 3x + 9 .
4
y = 2x 2 3x + 9
3
3
= 2 3 + 9
4
4
1
= 10
8
3
1
The vertex is ,10 4
8
2
4
PAS5.3.4 Coordinate geometry
d
For the x-intercept, y = 0.
y = 2x 2 3x + 9
= 2(0)2 3(0) + 9
=9
The curve crosses the y-axis at y = 9 . (Or simply look for the
constant term.)
For the y-intercept, x = 0 .
y = 2x 2 3x + 9
2x 2 3x + 9 = 0
(x + 3)(3 2x) = 0
x + 3 = 0or3 2x = 0
1
x = 3x = 1
2
The curve crosses the x-axis at –3 and 1.5.
y
e
10
9
8
7
6
5
4
3
2
1
–3
–2
–1
0
–1
1
2 x
–2
Part 1
Parabolas
5