Question

factor to solve quadtatic equations

1. what are the coordinates of the vertex of the graph? is it maximum or ninimum?
A- (1,0); minimum
B- (0, 1); maximum
C- (0,1); minimum
D- (1,0); maximum

2.graph the function and identify the domain and range.
y=-0.5x^2

A- domain: (♾, ♾) range: [0, ♾)
B- domain: (♾, ♾) range: (-♾, 0]
C- domain: (-♾, ♾) range: [0, ♾)
D- domain: (-♾, ♾) range: (-♾, 0]

3. how is the graph of y=-2x^2-4 different from the graph of y=-2x^2?

A- it is shifted 4 units to the left
B- it is shifted 4 units up
C- it is shifted 4 units to the right
D- it is shifted 4 units down

4. graph the function. identify the vertex and axis of symmetry.
f(x)=-x^2-x+2

i don't know, i got D

5. graph the function. identify the vertex and axis of symmetry.
f(x)=-2x^2+4x+1

i don't know, i got D

6. what are the solutions of the equation x^2-4=0? use the graph of the related function

A- there are two solutions: -2 and 2.
B- there are two solutions +_ square root of 2
C- there are no real number solutions.
D- there are two solutions: -2 and 2.

7. solve the equation using square roots.
5x^2-45=0

A- -3, 3
B- -9, 9
C- -square root of 3, square root of 3
D- no real number solutions

8. solve the equation using the zero product property.
(x-9)(x+7)=0

A- 9, 7
B- -9, -7
C- -1, 1
D- 9, -7

9. for questions 9-10, what are the solutions of the equation?
z^2-6z-27=0

A- 3, 9
B- 3, -9
C- -3, 9
D- -3, -9

10. c^2-4c=0

A- 0, -4
B- 0, square root of 4
C- 0- 4
D- 1, -square root of 4

Answer 1

1. B
2. D
3. D
4. A
5. B
6. D
7. A
8. D
9. C
10. C

Answer 2

Here are some ideas that should help you get started.

#1 - no idea, since no graph

#2 - there are lots of online graphing sites which can help you. The vertex is at (0,0), so that should help you determine the range. The domain of ALL polynomials is (-∞,∞)

#3 - well, 4 is subtracted from each y-value, so the graph is shifted down by 4

#4 - I don't know how you got D, since you showed no choices. But
-x^2-x+2 = -(x+2)(x-1)
The vertex is always on the axis of symmetry, midway between the roots. In this case, that is at x = -1/2. So find y(-1/2) to locate the vertex.

#5 - use the quadratic formula to find the roots, as in #4, or recall that for y=ax^2+bx+c, the vertex lies at x = -b/2a

#6 - x^2-4 = (x-2)(x+2)

#7 5x^2-45=0
x^2 - 9 = 0
(x-3)(x+3) = 0
#8 the zero product property states that if ab=0 then either a=0 or b=0
So, you have either
x-9 = 0
or
x+7 = 0

#9 z^2-6z-27=0
(z-9)(z+3) = 0

#10 c(c-4) = 0

Answer 3

Anonymous is 100% correct

Answer 4

oobleck yess, get it☝