1
00:00:01,280 --> 00:00:02,880
NARRATOR: Exercise three.

2
00:00:02,880 --> 00:00:08,400
A parabola has vertex (1, 3)
and passes through the point (3, 11).

3
00:00:08,400 --> 00:00:10,160
Find its equation.

4
00:00:10,160 --> 00:00:13,760
We know that an equation
given in turning point form -

5
00:00:13,760 --> 00:00:18,720
that is in the form
y = a(x – h) all squared plus k -

6
00:00:18,720 --> 00:00:21,560
has a vertex at (h, k).

7
00:00:21,560 --> 00:00:24,880
So in this case,
with the vertex at (1, 3),

8
00:00:24,880 --> 00:00:32,520
the equation will be of the form
y = a(x – 1) all squared, + 3.

9
00:00:32,520 --> 00:00:35,120
So with the equation in this form,

10
00:00:35,120 --> 00:00:39,840
the point (3, 11) can be used
to calculate a.

11
00:00:39,840 --> 00:00:42,680
So we substitute
the point (3, 11) in -

12
00:00:42,680 --> 00:00:46,320
that is, when x = 3, y = 11.

13
00:00:46,320 --> 00:00:51,800
In simplifying and solving for a,
we find that a is equal to 2.

14
00:00:53,160 --> 00:00:55,720
So therefore the equation
of the parabola

15
00:00:55,720 --> 00:00:59,400
which has a vertex at (1, 3) and
passes through the point (3, 11)

16
00:00:59,400 --> 00:01:04,400
is y = 2(x-1) all squared, + 3.