Chapter Review - Nelson Education

Since I multiplied all the
y = 2(x - 3) 2
a = 2, y x 2 6 y 2x 2 the graph of y = x 2 vertically by
x 1 2 3 4 5
y-coordinates of the points on the
red graph by 2. The vertex stays at
y 8 2 0 2 8
(3, 0). The equation of this graph
is y = 2(x - 3) 2 .
y 2(x 3) 2 y (x 3) 2
y
To draw this new blue graph,
8
I applied a vertical stretch by
6
a factor of 2 to the red graph.
4
The blue graph looks correct
because the graph with the
2
greater a value should be
x
0
narrower than the other graph.
-4 -2 2 4 6
-2
-4
y = 2(x - 3) 2 - 8
I knew that k = -8. I subtracted
8 from the y-coordinate of each
x 1 2 3 4 5
point on the blue graph. The
y 0 6 8 6 0
vertex is now (3, 8). The
equation of the graph is
y = 2(x - 3) 2 - 8.
y 2(x 3) 2 8
y
4
y 2(x 3) 2
2
Since k 6 0, I knew that I had to
x
translate the blue graph 8 units
-4
0
-2 2 4 6
down to get the final black graph.
-2
-4
-6
-8
Kevin’s Solution: Applying a vertical stretch first
8
4
y
Since a 2, I decided to stretch
a factor of 2. I multiplied the
2
y-coordinate of each point on
x
the graph of y = x 2 by 2.
-4
0
-2 2 4 6
The equation of the resulting red
-2
graph is y = 2x 2 .
-4
264 5.3 Graphing Quadratics in Vertex Form Draft Evaluation Copy
NEL