1. Given: h(x) = 6 x2 - x – 10 find h(-3)
You must square first.
This is called evaluating using DIRECT SUBSTITUTION.

2. Given: h(x) = 6 x2 - x – 10 find h(-3)
We can also evaluate a function using
SYNTHETIC SUBSTITUTION.
Given: h(x) = 6 x2 - x – 10 find h(-3)
Step 1: We need the coefficients: 6, -1, -10
Step 2:
-18 57
Solution 6
-19 47
Step 3: Bring down the 6. Multiply it by -3 and then add it to the – 1.

3. Given: h(x) = 4 x3 - x – 10 find h(2)
Step 1: We need the coefficients: 4, 0, -1, -10
Step 2: 8 16 30 4 8 15 20
Given: h(x) = 4 x4 - x2 – 10 We’d have to replace all the “missing” x’s with zeros…
h(x) = 4x4 + 0x3 – x2 + 0x - 10

4. How could you find it’s x intercepts?
Consider: h(x) = - x2 - x + 10
What’s it’s
y intercept?
Does it flip up or down?
What does it graph?
Vertex?
How could you find it’s x intercepts?

5. Consider: h(x) = - x2 - x + 10
It’s called an EVEN function because it is of degree: 2
It’s flips down because the leading coefficient is -1

6. Consider: h(x) = 3x2 - x + 10
It’s called an EVEN function because it is of degree: 2
It’s flips UP because the leading coefficient is 3

7. Consider: h(x) = - x3 - x + 10
It’s called an ODD function because it is of degree: 3
It’s flips down because the leading coefficient is -1

8. Consider: h(x) = 2x3 - x + 10
It’s called an ODD function because it is of degree: 3
It’s flips UP because the leading coefficient is 2