Download Test Unit 2 Answers - hhs

K/U
APP
COMM
TIPS
TEST Unit 2—Chapters 2 and 3
NAME : __________________________________________
PART A: Multiple Choice
Place the upper case letter corresponding to the correct answer on the blank to the left of the
question. [10] K/U
____
____
1. Expand and simplify
a.
b.
.
c.
d.
3. Which shows the polynomial fully factored?
a.
b.
____
6 x2  8
2. Which shows the polynomial fully factored?
a.
b.
____
c.
d.
c.
d.
4. Factor the quadratic equation fully.
a.
b.
c.
d.
____
5. Which shows the factorizations of the polynomial
a.
c.
b.
d.
?
____
6. What is the y-intercept of the quadratic function
?
The y intercept is the “C” term. Multiply the numbers without the xs: 4 x -3 = -12
a. (0, 12)
c. (0, 2)
b. (0, 3)
d. (0, 12)
____
? zeros at ½ and 5... must be answer (a)
7. Which is the graph of the function
y
a.
–10 –8
–6
–4
20
8
16
6
12
4
8
2
4
–2
–2
2
4
6
8
10
x
–4
–6
–12
–8
–16
–10
–20
10
15
8
12
6
9
4
6
2
3
–2
–2
2
4
6
8
10
x
–15 –12 –9
–3
–3
–6
–6
–9
–8
–12
–10
–15
____
9. Solve.
a. x = 12, 0
–6
–4
8. What is not another name for the root of an equation?
a. zeros
c. vertex
b. x-intercepts
d. Solutions
____ 10. Solve by factoring.
4
8
12
16
20
x
3
6
9
12
15
x
y
d.
____
a. x = 2, 7
b. x = 2, 7
–4
–4
–8
y
–6
–20 –16 –12 –8
–4
b.
–10 –8
y
c.
10
c. x = 2, 7
d. x = 2, 7
x2 – 12x = 0
...
c. x = 12, 0
x( x – 12) = 0
...
x = 0 or x = 12
PART B:Full Solution
Place your full solutions, showing your work, in the space provided.
1) Expand and simplify. [3] K/U
(n  2)(3n  7)  2n(2n  3)
3n2 + 7n – 6n – 14 + 4n2 + 6n
=
3n2 + 4n2 + 7n – 6n + 6n – 14
= 7n2 + 7n - 14
2) Factor completely. [3] K/U
 6 x 2  15 x  9
-3( 2x2 + 5x + 3)
=
-3 (2x2 + 3x + 2x + 3)
=
-3(x (2x + 3) + 1 (2x + 3))
=
-3 (x +1) (2x+3)
3) Can the following expression be factored further? If so, factor completely. [2] K/U
( x 2  4)( x 2  4)
(x2+4) (x -2)(x+2)
4) Express the following quadratic function in standard form. [2] K/U
f ( x)  ( x  2)( x  4)
f(x) = - (x2 -2x -4x + 8)
f(x) = -x2 + 6x - 8
5) For the quadratic function below, determine the coordinates of the vertex, without graphing.
Show your work. You may give your answer as a fraction in lowest terms or as a decimal. [4]
K/U
g ( x)  (2 x  1)( x  2)
Find zeros:
2x + 1 =0
x-2 = 0
x = -1/2
x=2
Vertex is at
x = (-1/2 + 2)/2
x = ¾ (or 0.75)
plug back in:
g(0.75) = (2 (0.75) + 1)(0.75 – 2)
g(0.75) = (2.5)(-1.25)
= -3.125
Final answer: Vertex (0.75, -3.125)
6) Use factoring to solve the following quadratic equations. [8] APP
a) x 2  x  12  0
b) 2 x 2  5 x  0
(x +4)(x-3) = 0
x = -4 or x = 3
x (2x – 5) = 0
x=0
x = 0 or
c) 6 x 2  7 x  5
6x2 - 7x – 5 = 0
6x2 - 10x + 3x – 5 = 0
2x (3x – 5) + (3x – 5) = 0
(2x + 1)(3x – 5) = 0
x = -1/2
or x = 5/3
or
2x -5 = 0
x = 5/2
7) John hits a golf ball into the ocean from a cliff that is 90 metres high. The function
h(t )  5t 2  15t  90 gives the height of the golf ball above the water, where h(t) is the height
in metres and t is the time in seconds. When will the ball hit the water? You must show an
algebraic solution using what we have learned in this unit. [4] APP
When the ball hits the water, height is zero (h(t) = 0). Solve for time t.
h(t )  5t 2  15t  90
0 = -5t2 – 15t + 90
0 = -5 (t2 + 3t – 18)
notice the signs. Be careful!
0 = -5 (t2 + 6t – 3t – 18)
0
= -5(t (t + 6) – 3(t + 6))
0
= -5(t-3)(t+6)
Solutions:
(t-3) = 0
or
t+6 = 0
t=3
or
t = -6
Only the positive time is possible. So final answer, the ball will hit the water after 3 seconds.
8)
Given the following quadratic function, [8] APP
y  x 2  2 x  15
a) Determine the y-intercept.
In standard form, the y intercept is the C term: y = -15
b) Find the zeros.
Factor:
y = (x + 5)(x – 3)
zeros at x = -5
and
x=3
c) Find the axis of symmetry.
Axis of symmetry is between the zeros. Take the average:
x = (-5 +3)/2
x = -1
don’t forget the “x =”
d) Find the vertex.
Vertex is along the axis of symmetry. Plug x = -1 into function.
y = (-1)2 + 2(-1) – 15
y = 1 – 2 – 15
y = -16
So the vertex is at V(-1,-16)
e) Graph the function below, showing your results from the other parts of this question clearly.
Not quite right, but as close as I can graph it using the computer :)
Zeros at x = 3, x = -5
Vertex V(-1,16)
1
4
The quadratic function y  2 x 2  5x  3 has axis of symmetry x =  1 . Will the
9)
function have a maximum or a minimum value? What is the maximum or minimum value? You
must give your answer as a fraction in lowest terms. [3] APP
The parabola is opening up (A term is positive), so it will have a minimum value, and no
maximum (goes upwards to infinity).
sub in x = -1 ¼
(this is equal to -5/4)
y = 2 (-5/4)2 + 5(-5/4) – 3
y = 2 (25/16) - 25/4 – 3
y = 25/8 - 25/4 – 3
y = 25/8 – 50/8 - 3
y = -25/8 – 3
y = -25/8 – 24/8
y = -49/8
The minimum value is -49/8
10)
What values of “b” will allow the following to be factored? You must show your
logic. Hint: You should be getting eight solutions to this question. [5] TIPS
x 2  bx  24
To factor, must find a number that multiplies to -24 and adds to b.
The numbers that multiply to -24 are:
1x-24
-1 x 24
2 x -12
-2 x 12
3 x -8
-3 x 8
4 x -6
-4 x 6
so
b = -23
b = +23
b = -10
b = +10
b = -5
b = +5
b = -2
b=2
So the values of b that work are
(since b = 1 – 24)
b = 2, -2, 5, -5, 10, -10, 23, -23