QUESTION IMAGE
Question
question #5
george and gwen are determining the point(s) of discontinuity by factoring and simplifying the following rational equation. determine who has an error in their work and what the correct answer is.
$y = \frac{2x^{2}+11x + 15}{x^{2}+8x + 15}$
| georges work | gwens work |
|---|
|step 1
factor numerator|$\frac{(2x + 5)(x + 3)}{x^{2}+8x + 15}$|step 1
factor numerator|$\frac{(2x + 5)(x + 3)}{x^{2}+8x + 15}$|
|step 2
factor denominator|$\frac{(2x + 5)(x + 3)}{(x + 5)(x + 3)}$|step 2
factor denominator|$\frac{(2x + 5)(x + 3)}{(x + 5)(x + 3)}$|
|step 3
simplify common factor(s)|$\frac{(2x + 5)}{(x + 5)}$|step 3
simplify common factor(s)|$\frac{(2x + 5)}{(x + 5)}$|
|step 4
determine the point of discontinuity|$(3,\frac{11}{8})$|step 4
determine the point of discontinuity form|$(-3,-\frac{1}{2})$|
gwen determined the point of discontinuity incorrectly. the ordered pair should be $(5,\frac{3}{2})$
george determined the point of discontinuity incorrectly. the ordered pair should be $(-3,-\frac{1}{2})$
gwen determined the point of discontinuity incorrectly. the ordered pair should be $(3,\frac{11}{8})$
george determined the point of discontinuity incorrectly. the ordered pair should be $(-5,0)$
Step1: Factor the numerator and denominator
For the numerator \(2x^{2}+11x + 15\), we find \(2x^{2}+11x + 15=(2x + 5)(x+3)\) using factoring by grouping. For the denominator \(x^{2}+8x + 15=(x + 5)(x + 3)\). So the rational - function becomes \(y=\frac{(2x + 5)(x + 3)}{(x + 5)(x + 3)}\).
Step2: Simplify the rational function
Cancel out the common factor \((x + 3)\) (where \(x
eq - 3\)), and we get \(y=\frac{2x + 5}{x + 5}\), \(x
eq-3\).
Step3: Determine the point of discontinuity
A point of discontinuity occurs when the original function is undefined but the simplified function is defined. The original function \(y=\frac{2x^{2}+11x + 15}{x^{2}+8x + 15}\) is undefined when \(x^{2}+8x + 15 = 0\), i.e., \((x + 5)(x + 3)=0\), so \(x=-5\) or \(x=-3\). After simplification, the function \(y=\frac{2x + 5}{x + 5}\) is undefined at \(x=-5\) and has a removable discontinuity at \(x=-3\). To find the \(y\) - value of the removable discontinuity, substitute \(x=-3\) into the simplified function \(y=\frac{2x + 5}{x + 5}\), we get \(y=\frac{2\times(-3)+5}{-3 + 5}=\frac{-6 + 5}{2}=-\frac{1}{2}\).
George made an error in step 4. He incorrectly determined the point of discontinuity. The correct point of discontinuity (the removable one) is \((-3,-\frac{1}{2})\).
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George determined the point of discontinuity incorrectly. The ordered pair should be \((-3,-\frac{1}{2})\)