Classify the following functions as one of the types of functions that we have discussed.
Quadratic Function
Find an equation of the quadratic below:
Find a formula for a cubic function if (−2) = 𝑓(1) = 𝑓(3) = 0 and 𝑓(2) = 8
f(x) = -x^3 + 6x^2 – 13x + 8
Ex 1) Let f(x) = 2x – 3. Find
a) f(–5)
b) f(2)a) f(–5) = 2(-5) – 3
= -10 – 3
= -13
b) f(2) = 2(2) – 3
= 4 – 3
= 1
Ex 2) Use the vertical line test to identify graphs in which y is a function of x.
a) b) c) d)
a) Yes – The vertical line test passes.
b) No – The vertical line test fails.

c) Yes – The vertical line test passes.
d) No – The vertical line test fails.
Ex 3) Determine whether the relations are functions:
b) c)) 2𝑥 + 𝑦2 = 16 𝑦 = 𝑥2 + 1 𝑦 = 𝑐𝑜𝑠𝑥
a) This is a function because it produces a unique output for each input. The domain is all real numbers, and the range is all real numbers greater than or equal to 4.
b) This is a function because it produces a unique output for each input. The domain is all real numbers, and the range is all real numbers greater than or equal to 1.
c) This is a function because it produces a unique output for each input. The domain is all real numbers, and the range is all real numbers greater than or equal to -1.
Ex 4) Let and𝑓 ( ) = 𝑥+1 𝑥−2 𝑔 𝑥( ) = 𝑥 + 2
Find the domain of each function.
f(x) = (x+1)/(x-2)
Domain of f(x) = {x | x ≠ 2}
g(x) = x + 2
Domain of g(x) = All real numbers
Ex 5) Let f(x) = 2x – 3. State the domain of the function, and find:

a) f(–5)
b) f(2)
c) f(a +1)
d) f(x + h)
Domain: x ∈ R
a) f(–5) = 2(-5) – 3 = -13
b) f(2) = 2(2) – 3 = 1
c) f(a + 1) = 2(a + 1) – 3 = 2a + 1 – 3 = 2a – 2
d) f(x + h) = 2(x + h) – 3 = 2x + 2h – 3
Ex 6) Simplify the difference quotient for 𝑓 𝑥+ℎ( )−𝑓(𝑥)
ℎ 𝑓 ( ) = 3 − 𝑥2
Piecewise Functions:
𝑓 𝑥+ℎ( )−𝑓(𝑥) = (3 − (𝑥+ℎ)2) − (3 − 𝑥2)
= (3 − 𝑥2 − 2𝑥ℎ − ℎ2) − (3 − 𝑥2)
= −2𝑥ℎ − ℎ2
Simplified Difference Quotient:
𝑓 𝑥+ℎ( )−𝑓(𝑥) = −2𝑥ℎ − ℎ2

Ex 7) Graph the piecewise function 𝑓 ( ){𝑥2 − 2𝑥 + 1 𝑓𝑜𝑟 𝑥 < 2 𝑥 − 1| | 𝑓𝑜𝑟 𝑥 ≥2

For x < 2
f(x) = x2 – 2x + 1
f(0) = 0 – 0 + 1 = 1
f(1) = 1 – 2 + 1 = 0
f(2) = 4 – 4 + 1 = 1

For x ≥ 2
f(x) = x – 1
f(2) = 2 – 1 = 1
f(3) = 3 – 1 = 2
f(4) = 4 – 1 = 3