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elect the correct answer.Which graph is the graph of the function 𝑟⁡(𝑥)=𝑥2+6⁢𝑥+82⁢𝑥+8 ?

Question

Select the correct answer.

Which graph is the graph of the function

r(x)=x2+6x+82x+8 r(x) = \frac{x^2 + 6x + 8}{2x + 8} ?

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Solution

To determine which graph corresponds to the function r(x)=x2+6x+82x+8 r(x) = \frac{x^2 + 6x + 8}{2x + 8} , we need to analyze the function step by step.

Step 1: Simplify the Function

First, let's simplify the function if possible.

r(x)=x2+6x+82x+8 r(x) = \frac{x^2 + 6x + 8}{2x + 8}

Factor the numerator:

x2+6x+8=(x+2)(x+4) x^2 + 6x + 8 = (x + 2)(x + 4)

So the function becomes:

r(x)=(x+2)(x+4)2(x+4) r(x) = \frac{(x + 2)(x + 4)}{2(x + 4)}

Step 2: Cancel Common Factors

We can cancel the common factor (x+4)(x + 4) in the numerator and the denominator, but we must note that x4 x \neq -4 because it would make the denominator zero.

r(x)=x+22forx4 r(x) = \frac{x + 2}{2} \quad \text{for} \quad x \neq -4

Step 3: Identify the Simplified Function

The simplified function is:

r(x)=x+22 r(x) = \frac{x + 2}{2}

This is a linear function with a slope of 12\frac{1}{2} and a y-intercept of 1.

Step 4: Consider the Domain

The function r(x) r(x) is defined for all x x except x=4 x = -4 . At x=4 x = -4 , there is a vertical asymptote or a hole in the graph.

Step 5: Sketch the Graph

  1. Linear Part: The graph of r(x)=x+22 r(x) = \frac{x + 2}{2} is a straight line with a slope of 12\frac{1}{2} and a y-intercept of 1.
  2. Discontinuity: There is a hole at x=4 x = -4 .

Step 6: Match with Given Graphs

To match the function with the correct graph, look for a graph that:

  • Is a straight line with a slope of 12\frac{1}{2}.
  • Has a y-intercept at y=1 y = 1 .
  • Has a hole at x=4 x = -4 .

Conclusion

The correct graph will be a straight line with the specified slope and y-intercept, and it will have a hole at x=4 x = -4 .

Without the actual graphs to compare, this is the detailed process to identify the correct graph. If you have multiple-choice options, look for the graph that fits these criteria.

This problem has been solved

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