The function is defined for all real numbers such that . When the graph of is drawn in the -plane, it appears as a straight line with a single hole (a point that is missing from the graph). What is the -coordinate of that hole?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 53 | Free SAT Prep | Aniko

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Question 53·Medium·Nonlinear Functions

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$g(x) = \dfrac{x^{2}-9}{x-3}$

The function $g$ is defined for all real numbers $x$ such that $x \ne 3$ . When the graph of $y = g(x)$ is drawn in the $xy$ -plane, it appears as a straight line with a single hole (a point that is missing from the graph). What is the $x$ -coordinate of that hole?

Your answer

(Express the answer as an integer)

For rational-function graph questions, first factor the numerator and denominator and cancel any common factors to see the "underlying" simple graph (often a line or a simple curve). Then, always go back to the original denominator to find where it is zero—those $x$ -values are excluded from the domain. If a factor causing a zero in the denominator cancels, the graph has a hole there (same shape as the simplified graph but missing one point); if it does not cancel, you have a vertical asymptote. This approach lets you quickly identify holes and asymptotes without plotting many points.

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Look at the form of the function

Rewrite $g(x)$ by factoring the numerator $x^2 - 9$ . How does that help you see the shape of the graph?

Think about domain restrictions

For what $x$ -value does the denominator become zero? What does that mean for the function's domain and graph?

Compare the simplified graph to the original function

After canceling any common factors, you get the equation of a simple line. How does the original domain restriction show up on the graph of that line?

Desmos Guide

Graph the original function

In Desmos, enter y = (x^2 - 9)/(x - 3) as one expression. Look carefully near the $x$ -value where the denominator would be zero to see how the graph behaves.

Compare with the simplified line

On a new line, enter y = x + 3. You should see that this line lies exactly on top of the graph of the rational function everywhere it is defined, except at one missing point on the rational graph.

Find the x-value of the hole

Use the graph or a table (click on the graph of the rational function and open the table) to find the $x$ -value where the rational expression is undefined while the line y = x + 3 is defined. That $x$ -value is the coordinate of the hole.

Step-by-step Explanation

Factor and simplify the function

Start by factoring the numerator of $g(x)$ .

The numerator is a difference of squares:

x^2 - 9 = (x-3)(x+3)

g(x) = \frac{x^2 - 9}{x - 3} = \frac{(x-3)(x+3)}{x-3}

For $x \ne 3$ , the $(x-3)$ terms cancel, leaving

g(x) = x + 3 \quad \text{(for } x \ne 3\text{)}

This shows the graph looks like the straight line $y = x + 3$ , except possibly at the excluded value from the domain.

Connect the domain restriction to the graph

From the original definition, $g$ is defined only for $x \ne 3$ because the denominator $x - 3$ cannot be zero.

That means:

The simplified expression $y = x + 3$ would normally be defined for all real $x$ .
But the original function $g(x)$ is missing the point where $x = 3$ .

So the graph is the line $y = x + 3$ with exactly one point removed at the $x$ -value where the denominator was zero.

Identify the x-coordinate of the hole

A "hole" (removable discontinuity) in a rational function happens at an $x$ -value where a factor cancels but the original function is still undefined.

Here, the factor $x-3$ canceled, and the original denominator is zero at $x=3$ , so $g(x)$ is undefined at $x=3$ even though $y = x + 3$ is defined there.

Therefore, the $x$ -coordinate of the hole is $3$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation