Originally Posted by

**juiicycouture** ...

1. Solve the following equation:

(x+1)^2 - 3(x+1) = 0

Factor the term:

(x+1)^2 - 3(x+1) = (x+1)((x+1) - 3) = (x+1)(x-2) = 0

Therefore the solutions are: x = -1 or x = 2

Originally Posted by

**juiicycouture** .

2. What is the maximum possible number of turning points for each of the following functions?

a) linear function

b) quadratic function

c) cubic function

d) quartic function

If n is the degree (or do you say grade?) of the function then the maximum number of turning points is n-1: Code:

prob degree max. number of turning points
-------------------------------------------
a) 1 0
b) 2 1 (= vertex)
c) 3 2
d) 4 3

Originally Posted by

**juiicycouture** 3. For the function f(x) = (x-4)(x+1)(x-1)

a) find the domain and range

b) find the real zeros, to the nearest tenth necessary

c) the y-intercepts

d) the end behaviour (what is this?)

e) any symmetry

f) the number of turning points

a) There aren't any restrictions for x thus d = IR (the set of real numbers).

Because f has the degree 3 and the coefficient of x³ is 1 the values of the function reach from -∞ to +∞

b) f(x) = 0 if one of the factors is zero. Therefore the zeros are: x = -1, x = 1, x = 4

c) The y-intercept is f(0) = (-4)(1)(-1) = 4

d) If x --> -∞ then f(x) --> -∞

If x --> ∞ then f(x) --> ∞

e) Because f has the degree 3 there is only possible a point inflection. The centre of symmetry could only be the inflection point. With your function the graph of f is symmetric to the point ((4/3), -(56/27))

f) see #2: thus 2

EB