Question

mpm2d0
quadratic graphs
name: salma bouaj
multiple choice: circle the correct answer.

1. the graph of the relation ( y = 5(x + 2)^2 ) is a parabola. which way does this parabola open?

a) up
b) right
c) left
d) down

1. what are the coordinates of the vertex of the relation ( y = (x - 2)^2 + 4 )?

a) ( (2, 4) )
b) ( (2, -4) )
c) ( (-2, 4) )
d) ( (-2, -4) )

1. what is the equation of the axis of symmetry for the relation ( y = (x - 4)^2 - 3 )?

a) ( x = 3 )
b) ( x = -3 )
c) ( x = 4 )
d) ( x = -4 )

1. the graph of ( y = x^2 ) opens down, is vertically stretched by a factor of 2, and then translated 2 units to the right and 11 units up. what is the equation of the new parabola?

a) ( y = -2(x - 2)^2 + 11 )
b) ( y = -2(x + 2)^2 + 11 )
c) ( y = -2(x - 2)^2 - 11 )
d) ( y = 2(x - 2)^2 + 11 )

1. what transformation will move the parabola ( y = (x + 5)^2 - 3 ) to the position modeled by ( y = (x - 2)^2 + 2 )?

a) 5 units to the left, 3 units up
b) 5 units to the right, 3 units down
c) 7 units to the right, 5 units up
d) 7 units to the left, 5 units down
use the space below to answer the following questions. show your work.

1. determine if the following tables represent linear or quadratic relationships using differences in y - values.

a)

x	y
-1	2
0	1
1	2
2	5

b)

x	y
-1	1
0	4
1	7
2	10

graph
a) ( y = (x - 2)^2 - 4 )
b) ( y = -\frac{1}{2}(x + 1)^2 + 3 )

Explanation:

Let's tackle problem 6a first (determine if the table represents linear or quadratic using differences in y - values).

Step 1: List the x and y values

For part a), the x - values are \(-2,-1,0,1,2\) and the corresponding y - values are \(5,2,1,2,5\).

Step 2: Find the first differences (differences between consecutive y - values)

• Between \(x=-2\) ( \(y = 5\)) and \(x=-1\) ( \(y = 2\)): \(2 - 5=-3\)
• Between \(x=-1\) ( \(y = 2\)) and \(x = 0\) ( \(y = 1\)): \(1 - 2=-1\)
• Between \(x = 0\) ( \(y = 1\)) and \(x = 1\) ( \(y = 2\)): \(2 - 1 = 1\)
• Between \(x = 1\) ( \(y = 2\)) and \(x = 2\) ( \(y = 5\)): \(5 - 2 = 3\)

Step 3: Find the second differences (differences between consecutive first differences)

• Between \(-3\) and \(-1\): \(-1-(-3)=2\)
• Between \(-1\) and \(1\): \(1 - (-1)=2\)
• Between \(1\) and \(3\): \(3 - 1 = 2\)

Since the second differences are constant (all equal to 2), the relationship is quadratic.

Now for problem 6b:

Step 1: List the x and y values

For part b), the x - values are \(-2,-1,0,1,2\) and the corresponding y - values are \(2,3,4,5,6\).

Step 2: Find the first differences (differences between consecutive y - values)

• Between \(x=-2\) ( \(y = 2\)) and \(x=-1\) ( \(y = 3\)): \(3 - 2 = 1\)
• Between \(x=-1\) ( \(y = 3\)) and \(x = 0\) ( \(y = 4\)): \(4 - 3 = 1\)
• Between \(x = 0\) ( \(y = 4\)) and \(x = 1\) ( \(y = 5\)): \(5 - 4 = 1\)
• Between \(x = 1\) ( \(y = 5\)) and \(x = 2\) ( \(y = 6\)): \(6 - 5 = 1\)

Since the first differences are constant (all equal to 1), the relationship is linear.

Answer:

• For 6a: The relationship is quadratic (because the second differences of the y - values are constant).
• For 6b: The relationship is linear (because the first differences of the y - values are constant).