Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
1. Quadratic Graphs and Their Properties
Continue to next subchapter

Exercise 47 Page 552

Practice makes perfect
a
Let's start by simplifying the exponent in the second equation.
y=(ax)^2
y=a^2x^2

We will be comparing Equation One, y_1=ax^2, and Equation Two, y_2=a^2x^2. Equation One:& y_1=ax^2 Equation Two:& y_2=a^2x^2 The difference between the equations is that in Equation Two, it is a^2 instead of just a. Recall that when we square a number, the result is always positive. Therefore, Equation Two will be always be positive, regardless of the a-value, because we are multiplying two positive numbers. Ifa is positive or negative: y_2=a^2x^2 ⇒ y_2=(+)(+) ⇒ y_2=(+) In Equation One, if a is positive we will multiply two positive numbers together, resulting in the function being positive. Ifa is positive: y_1=ax^2 ⇒ y_1=(+)(+) ⇒ y_1=(+) However, when a is negative we will be multiplying a negative value by the always-positive x^2 value. Therefore, when a is negative, the function will be negative. Ifa is negative: y_1=ax^2 ⇒ y_1=(-)(+) ⇒ y_1=(-) In conclusion, this all tells us that when a is positive, both Equation One and Two will be positive. However, when a is negative Equation One will be negative and Equation Two will be positive. Thus, the graphs will be in the same quadrant when a is positive.

b
We want to see which parabola will be wider. Recall that wider parabolas have smaller |a|-values. Since we want to know when y_1=ax^2 will be wider than y_2=a^2x^2, we want to find when |a| will be smaller than |a^2|. Note that a^2 is always positive, so we do not need absolute value signs around it. |a|

For every |a|-value greater than one, we can see that |a|

|a| a^2
2 4
3 9
4 16

We can see that when |a| is greater than one, |a|less than or equal to one and greater than zero. That is, 0<|a|≤ 1. Let's test some values to see what happens.

|a| a^2
1 1^2=1
1/2 1^2/2^2=1/4
1/4 1^2/4^2=1/16

We see that when 0<|a|≤ 1, |a| is not less than than a^2. Therefore, this will not be part of our solution. |a|1 In summation, we have found that |a|1. This means that the graph of y_1=ax^2 will be wider than y_2=a^2x^2 when |a|>1.