2. Flashback 10-3-12 In the figure below, <H <F; E, G, and I are collinear; and G is the midpoint of FH. To prove that HI FE given the conditions stated above, which of the following is a logical order for the 5 steps in the table below? A. 1, 2, 3, 4, 5 B. 1, 2, 3, 5, 4 C. 1, 2, 4, 3, 5 D. 1, 4, 2, 3, 5 E. 1, 5, 4, 2, 3 StatementReason 1. HG FGThe midpoint of a line segment divides the segment into 2 congruent segments 2. EGF IGHVertical angles are congruent 3. GHI GFEAngle-side-angle congruence theorem 4. EGF and IGH are vertical anglesDefinition of vertical angles 5. HI FECorresponding parts of congruent triangles are congruent

3. Joke of the day He said I was average -

4. but he was just being mean.

5. Finding domain Really finding what cannot be in the domain. Ex. : f(x) = √(x + 3) Think: What values would cause problems for us, assuming real number values? Square root functions cannot have So, And the domain is

6. Try this one Domain of g(x) = (√x)/ (x – 5)

7. How about this? A(s) = (√3 /4) s 2 where A(s) is the area of an equilateral triangle with side s.

8. Range F(x) = 2/x What do you expect it to be? Graph it. What does the range appear to be? How can we confirm?

9. Continuity A function is continuous if it has no gaps in the graph.

10. Discontinuous functions Removable discontinuity Jump discontinuity Infinite discontinuity

11. HW: p. 102 Ex. 5-12, 21-24

12. Exit Slip p. 104 #84