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Chapter 4 Review Package
Name: _________________
Breaking a quadrilateral down into its coefficient pieces: y = ax 2 + bx + c
a > 0 (+) parabola opens up;
a < 0 (-) parabola opens down
c is the y-intercept of the parabola
parabola has a min value
parabola has a max value
x-intercepts are the zeros (see them later)
Vertex formula: (x, y) =( -b / 2a , -(b2 – 4ac)/(4a) ) (vertex y-value is the min or the max)
(Use max or min part of 2nd TRACE in calculator to let the calculator find it)
Line of Symmetry: x = -b/2a
Domain: X is all real numbers
Range: Y is ≥ min if it opens up (a > 0)
Y is ≤ max if it opens down (a < 0)
1. Find for y = -x2 - 5x + 14
a. Opens: Up
Down
b. Domain: {x |
c. Range: {y |
d. Vertex: (
,
)
e. Line of Symmetry: x =
f. Y-intercept: (0,
)
2. Find for y = x2 + 13x + 40
a. Opens: Up
Down
b. Domain: {x |
c. Range: {y |
d. Vertex: (
,
)
e. Line of Symmetry: x =
f. Y-intercept: (0,
)
3. Find for y = -2x2 + 2x + 40
a. Opens: Up
Down
b. Domain: {x |
c. Range: {y |
d. Vertex: (
,
)
e. Line of Symmetry: x =
f. Y-intercept: (0,
)
4. Find for y = x2 – 6x + 9
a. Opens: Up
Down
b. Domain: {x |
c. Range: {y |
d. Vertex: (
,
)
e. Line of Symmetry: x =
f. Y-intercept: (0,
)
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Chapter 4 Review Package
Name: _________________
Finding zeros of polynomials and solving equations = 0 are the same thing. They can
be found be graphing the polynomial and finding the x-intercepts (2nd Trace, Zeros)
Examples:
f(x) = x(x + 4) (x – 1)
0 = (2x – 1)(3x + 2)
zeros of f(x) are when x = 0, x = -4 and x = 1
zeros of g(x) are when x = ½
and x = -2/3
x = 0, (x+4) = 0, (x – 1) = 0
x = -4
x=1
(2x – 1) = 0,
2x = 1
x=½
(3x + 2) = 0
3x = -2
x = -2/3
Find the zeros of
1. f(x) = (x – 2) (x + 3)
2. g(x) = (x + 6)(x – 1)
3. h(x) = (2x + 1)( x – 7)
4. j(x) = (3x + 4)(2x – 5)
5. f(t) = (t – 6)((t – 1)(t + 3)
Find the solutions to
1. (x + 2) (x + 5) = 0
2. (h – 2)(2h – 3) = 0
3. (x + 4)(x – 1) = 0
4. (m – 3)( 2m – 7) = 0
5. t(t + 4)(t – 8) = 0
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Chapter 4 Review Package
Name: _________________
Quadratic Regression: y = ax2 + bx + c
Put x-values in L1 and y-values in L2. Go to STAT – CALC and select 5: QuadReg
Write down values calculator gives for a, b and c for quadratic equation above.
Use new equation to predict values inside bounds of data (interpolation) or to predict
values beyond the bounds of the data (extrapolation).
Graph data using StatPlot (Zoom 9 sets up the window properly) and graph the regression
model using Y1 = ax2 + bx + c
X
0
2
4
6
8
9
y
42
20
15
25
50
61
Write the function: y = __________________________________
What would the model predict if x = 5?
Quadratic Equation:
Must be in the form of ax2 + bx + c = 0
-b  √b2 – 4ac
x-intercepts (zeros of the function) = ----------------------------2a
Remember a negative number squared is now a positive number!
Using the Quadratic Equation on the following:
1) x2 + 5x – 14 = 0
a= ______ b=______ c=______
x1 = _____ x2 = _____
2) 2x2 + 13x + 21 = 0
a= ______ b=______ c=______
x1 = _____ x2 = _____
3) 6x2 + 5x – 5 = 0
a= ______ b=______ c=______
x1 = _____ x2 = _____
4) x2 + 3 = 0
a= ______ b=______ c=______
x1 = _____ x2 = _____
What does getting a negative number under the radical (in #4) mean on the graph of the
problem?
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Chapter 4 Review Package
Name: _________________
Factoring:
Factor Cases: Looking at the signs
1) + + + = ( + )( + )
2) + - + = ( - )( - )
3) + - = ( + ) ( - ) where the |negative| > |positive|
4) + + - = ( + ) ( - ) where the |positive| > |negative|
Example 1: x² - 5x + 6 = 0
identify coefficients:
a = 1, b = -5, c = 6
If a ≠ 1, then break a down into its factors or factor a out of each term
Case 2 applies. x² - 5x + 6 = (x - )(x - )
Break down c into its factors:
16 and 23
If a = 1 then what combinations of the factors of c add up to b
So solution is x² - 5x + 6 = (x - 2)(x - 3)
1, 2, 3, 6
-5 = -2 + -3
Zeros are at x = 2 and x = 3
Factor the following problems:
1. y = x2 + 5x + 6
2. y = x2 + 15x + 56
3. y = x2 - 7x + 12
4. y = x2 + 2x + 1
5. y = x2 - 5x - 14
6. y = x2 - 6x + 9
7. y = x2 + x - 6
8. y = x2 - 12x + 32
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