1.your friend runs up to you, scared that he is not ready for the upcoming quadratics test. to him study, you will create four different quadratic functions. then demonstrate to him how to rewrite each function as a group of factors, if possible. •the function f(x) is a difference of squares.
•the function g(x) is a sum of squares.
•the function h(x) is a perfect square trinomial.
•the function j(x) can only have a gcf factored out of it.

2.given the function k(x) = x2, compare and contrast how the application of a constant, c, affects the graph. the application of the constant must be discussed in the following manners: •k(x + c)
•k(x) + c
•k(cx)
•c • k(x)

3.explain the grouping method of factoring. describe a scenario when the grouping method would be preferred over other methods and provide an example of this type of problem.

4.graph one of your 2nd degree functions from question 1. identify which function you used and the key features of your graph. explain how to find them algebraically.

5.using your graph from question 4, describe if the average rate of change is increasing or decreasing, from left to right. justify your observations by comparing the slopes calculated between at least three different pairs of points.

For Question number 1:

The function f(x) is a difference of squares. Is exactly what its name says, you take 2 squared quantities like x² and z² or a square of a number like 36 or 25, then put a minus sign between them. So 

f(x) = x² - 49 =factors to=> (x-7)(x+7) or 
f(x) = 64 - x² =factors to=> (8+x)(8-x) or 
f(x) = x² - k² =factors to=> (x+k)(x-k) and so on 

this is factored by taking the root of each term and rewriting them as a sum and difference. 

The function g(x) is a sum of squares. This is the same as above but with a + sign in between 

g(x) = x² + 25 or 
g(x) = 81 + x² or 
g(x) = x² + p² and so on 

These terms do not factor. 

The function h(x) is a perfect square trinomial. 
A perfect square trinimial is created by taking x (at any power) and adding or subtracting it to a number or letter, then squaring the whole quantity, and as a last, optional step, you expand this quantity. 

h(x) = (x+3)² = x² + 6x +9 or 
h(x) = (2x-5)² = 4x² - 20x + 25 or 
h(x) = (x² - t)² = x^4 - 2tx² + t² and so on 

The function j(x) can only have a GCF factored out of it. 
Here the equation must have an x in it, the x can be any power, and only the coefficients of the x's can be factored out, and there may be one number only term that also factors with the coefficients, like 

j(x) = 14x² + 21x - 7 =factors to=> 7(2x² + 3x - 1) or 
j(x) = 10x - 25t + 15 =factors to=> 5(2x² - 5t + 3) or 
j(x) = 64x³ + 16x² - 8x + 24 =factors to=> 8(8x³ + 2x² - x + 3) or 
j(x) = 64x³ + 16x² - 8x =factors to=> 8x(8x² + 2x - 1) and so on

a. geometricb. g[h] = (3/4)·g[h-1]; g[1] = 384c. g[h] = 384·(3/4)^(h-1)d. g[12] ≈ 16.2183 gallonse. g[12] ≈ 64.8732 quarts

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt