McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
Preparing for Standardized Tests
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Exercise 3 Page 587

Rewrite the fractions so that they have a common denominator, then subtract.

D

Practice makes perfect
We are given the following function and asked to find its x-intercepts. f(x)=2/x-1-x+4/3 Let's start with recalling that an x-intercept is a point where the graph intersects the x-axis. The y-coordinate of this point is 0. Hence, to find the x-intercepts of the given function we can substitute f(x) with 0 and solve it for x. 0=2/x-1-x+4/3First, we will rewrite the right-hand side as one rational expression. In order to do this, we will expand the first fraction by 3 and the second one by x-1 so that they have a common denominator. 0=2( 3)/3(x-1)-(x+4)(x-1)/3(x-1) Now we are able to subtract the fractions.
0=2(3)/3(x-1)-(x+4)(x-1)/3(x-1)
Simplify right-hand side
0=2(3)-(x+4)(x-1)/3(x-1)
0=6-(x+4)(x-1)/3(x-1)
0=6-(x^2-x+4x-4)/3(x-1)
0=6-x^2+x-4x+4/3(x-1)
0=- x^2-3x+10/3(x-1)
The fraction can be equal to 0 only if its numerator is equal to 0. This means that we need to solve the following equation. - x^2-3x+10=0 Let's first change the signs by multiplying both sides of the equation by - 1. Then we will solve it by factoring.
- x^2-3x+10=0
Factor
x^2+ 3x-10=0
x^2+ 5x-2x-10=0
x(x+5)-2(x+5)=0
(x-2)(x+5)=0
According to the Zero Product Property, at least one of the parentheses should be equal to 0. lx-2=0 x+5=0 ⇒ lx=2 x=- 5 We got that the zeroes, or x-intercepts of the function are - 5 and 2. The answer is D.