1. M3D7 Have out: Bellwork: pencil, red pen, highlighter, GP notebook,
Identify the following for each equation, then graph the equations.
Bellwork:
a) degree: ____
d) zeros: ________
b) leading coefficient: ____
e) y–intercept: ______
c) endpoint behavior: ____ 1) 2)
total:

2. b) leading coefficient: ____ e) y–intercept: ______ (0, 0)1) 3
a) degree: ____
0, 2 –1
d) zeros: ________ –1
b) leading coefficient: ____
e) y–intercept: ______
(0, 0)
c) endpoint behavior: ____ y x

3. b) leading coefficient: ____ e) y–intercept: ______ (0, 3)2) 4
a) degree: ____
–1, 1, 2, –3
d) zeros: ________
b) leading coefficient: ____
e) y–intercept: ______
(0, 3)
c) endpoint behavior: ____ y x
total:

4. “Name That Polynomial”
Recall from Unit 4 when we could write the equation of a parabola given the vertex and another point on the curve.
For example, write the equation for a parabola that has a vertex of (–5, 6) and goes through the point (1, 2).
Solve for a.
y = a(x – h)2 + k
2 = a(6)2 + 6
Substitute the vertex (–5, 6).
y = a(x + 5)2 + 6
2 = 36a + 6 –6 –6
Substitute the other point.
–4 = 36a
2 = a(1 + 5)2 + 6 36 36

5. We will use a similar method when we write the equation of a polynomial function. All we need are the zeros and one other point on the curve.
Step #1: Identify the zeros
Zeros:
x = –2, 0, 2 a)
Step #2: Write the equation with factors y
x + 2 x
x – 2
Step #3: Substitute the extra point (1, – 1) for x and P(x) x
Step #4: Solve for a.
(1, –1)
Step #5: Write the final equation.

6. Try problem (c) on your own.
zeros:
x = –2, 0, 2
x + 2 x
x – 2 x
Stop for a second… what we wrote above is a degree 3 polynomial, but look at the graph, it’s even degree. What’s going on?
(–1, –2)
There’s a double zero.
We must square x.
Try problem (c) on your own.

7. c) zeros: x = –3, –1, 1, 2 x + 3 x + 1 x – 1 x – 2y
(–2, 9)
x + 3
x + 1
x – 1
x – 2 x
This is a good time to check your work with the graph.
Does the leading coefficient suggest that the graph will open down?
Does the equation indicate that this is an even degree polynomial as pictured above?

8. d) What kind of polynomial is this? Odd degree + leading coefficient
double zero
triple zero
However, there are only 2 noticeable zeros: x
x = –1, 2
x = –1 is a DOUBLE ZERO, but there is something special about x = 2.
(1, –2)
The graph looks “cubic” through this root, therefore, it’s a TRIPLE ZERO.
x + 1
x – 2

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