Pearson Algebra 1 Common Core, 2011
PA
Pearson Algebra 1 Common Core, 2011 View details
1. Adding and Subtracting Polynomials
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Exercise 49 Page 491

a To find the equations for p and q in slope-intercept form, we need to reference the given graph. Let's find the equation for line p first.
We can find the slope, m, by choosing two points on the graph and measuring the distance between them. Let's use (1,1) and (2,3). Measuring up we can see the vertical distance is 2. When we count horizontally, the distance is 1.

m=rise/run ⇒ m=2/1 We can identify b using the y-value of the y-intercept, -1.

We can use the same process to find the equation for line q. Then, we can substitute those values into slope-intercept form to find the equations for each line.

line m b y=mx+b
p(x) 2 -1 p(x)=2x-1
q(x) 1/2 3 q(x)=1/2x+3


b We need to find the difference between each line at a given x-value. Let's use our equations from Part A, to write an equation that subtracts one from the other. The result will be the difference. Be sure to include absolute value symbols, since the distance between the lines cannot be negative.
D(x)=|(2x-1)-(1/2x+3)| We can simplify this equation further using the rules for subtracting polynomials.
D(x)=|(2x-1)-(1/2x+3)|
D(x)=|(2x-1)+(-1/2x-3)|
D(x)=|2x-1-1/2x-3|
D(x)=|(2x-1/2x)+(-1-3)|
â–Ľ
Evaluate
D(x)=|(2x/1-1x/2)+(-1-3)|
D(x)=|(2x/1(2)-1x/2)+(-1-3)|
D(x)=|(4x/2-1x/2)+(-1-3)|
D(x)=|(3x/2)+(-4)|
D(x)=|3x/2-4|
D(x)=|3/2x-4|
c To find the value of x that makes D(x)=0, we can substitute that value into the equation we created in Part B. We can remove the absolute value symbols since we are setting the absolute value equal to 0.
D(x)=|3/2x-4|
0=3/2x-4
â–Ľ
Evaluate
4=3/2x
8=3x
8/3=x
x=8/3
d When considering how the x-value in Part C relates to the graph of these points, remember that D(x) is the distance between the two lines. Therefore, if at any point D(x)=0 the two lines are intersecting. The x-value is the location where the two lines intersect.