Answer :
Let's analyze the function [tex]\( h(x) = \frac{2x - 1}{x + 3} \)[/tex] step by step to determine which statements about its graph are true or false.
1. The graph crosses the [tex]\( y \)[/tex]-axis at [tex]\((0, 3)\)[/tex].
To find where the graph crosses the [tex]\( y \)[/tex]-axis, we need to evaluate the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ h(0) = \frac{2(0) - 1}{0 + 3} = \frac{-1}{3} = -\frac{1}{3}. \][/tex]
So, the graph crosses the [tex]\( y \)[/tex]-axis at [tex]\((0, -\frac{1}{3})\)[/tex], not [tex]\((0, 3)\)[/tex].
Answer: False
2. The graph crosses the [tex]\( x \)[/tex]-axis at [tex]\((-2, 0)\)[/tex].
To find where the graph crosses the [tex]\( x \)[/tex]-axis, we need to solve for [tex]\( x \)[/tex] when [tex]\( h(x) = 0 \)[/tex]:
[tex]\[ \frac{2x - 1}{x + 3} = 0. \][/tex]
This equation is zero when the numerator [tex]\(2x - 1 = 0\)[/tex]:
[tex]\[ 2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}. \][/tex]
So, the graph crosses the [tex]\( x \)[/tex]-axis at [tex]\((\frac{1}{2}, 0)\)[/tex], not [tex]\((-2, 0)\)[/tex].
Answer: False
3. The graph has a vertical asymptote at [tex]\( x = -3 \)[/tex].
A vertical asymptote occurs where the denominator is zero and the numerator is not zero. Setting the denominator equal to zero:
[tex]\[ x + 3 = 0 \implies x = -3. \][/tex]
Thus, there is a vertical asymptote at [tex]\( x = -3 \)[/tex].
Answer: True
4. The graph has a hole at [tex]\( (3, 0) \)[/tex].
A hole occurs in the graph when both the numerator and the denominator have a common factor that cancels out, leading to an undefined point. Here, [tex]\(h(x) = \frac{2x - 1}{x + 3}\)[/tex] does not have such a factor that cancels out. Specifically, substituting [tex]\( x = 3 \)[/tex] into [tex]\( h(x) \)[/tex]:
[tex]\[ h(3) = \frac{2(3) - 1}{3 + 3} = \frac{6 - 1}{6} = \frac{5}{6} \neq 0. \][/tex]
There is no hole at [tex]\( (3, 0) \)[/tex].
Answer: False
5. The graph has a horizontal asymptote at [tex]\( y = 2 \)[/tex].
To find the horizontal asymptote, we look at the end behavior of the function as [tex]\( x \)[/tex] approaches infinity. For rational functions, the horizontal asymptote is determined by the ratio of the leading coefficients of the numerator and the denominator:
[tex]\[ \lim_{{x \to \infty}} \frac{2x - 1}{x + 3} = \frac{2x}{x} = 2. \][/tex]
So, the horizontal asymptote is [tex]\( y = 2 \)[/tex].
Answer: True
6. The domain of the function does not include the values [tex]\( x = -3 \)[/tex] and [tex]\( x = 3 \)[/tex].
The domain of the function is determined by the values of [tex]\( x \)[/tex] for which the denominator is not zero. The denominator is zero at [tex]\( x = -3 \)[/tex], making [tex]\( x = -3 \)[/tex] not part of the domain. However, there is no issue at [tex]\( x = 3 \)[/tex]:
[tex]\[ x + 3 \neq 0 \implies x \neq -3. \][/tex]
The domain excludes [tex]\( x = -3 \)[/tex], but [tex]\( x = 3 \)[/tex] is included in the domain.
Answer: False
Summarizing:
\begin{tabular}{|l|l|l|}
\hline The graph crosses the [tex]\( y \)[/tex]-axis at [tex]$(0,3)$[/tex]. & true & \textbf{false} \\
\hline The graph crosses the [tex]\( x \)[/tex]-axis at [tex]$(-2,0)$[/tex]. & true & \textbf{false} \\
\hline The graph has a vertical asymptote at [tex]$x=-3$[/tex]. & \textbf{true} & false \\
\hline The graph has a hole at [tex]$(3,0)$[/tex]. & true & \textbf{false} \\
\hline The graph has a horizontal asymptote at [tex]$y=2$[/tex]. & \textbf{true} & false \\
\hline The domain of the function does not include the values [tex]$x=-3$[/tex] and [tex]$x=3$[/tex]. & true & \textbf{false} \\
\hline
\end{tabular}
1. The graph crosses the [tex]\( y \)[/tex]-axis at [tex]\((0, 3)\)[/tex].
To find where the graph crosses the [tex]\( y \)[/tex]-axis, we need to evaluate the function at [tex]\( x = 0 \)[/tex]:
[tex]\[ h(0) = \frac{2(0) - 1}{0 + 3} = \frac{-1}{3} = -\frac{1}{3}. \][/tex]
So, the graph crosses the [tex]\( y \)[/tex]-axis at [tex]\((0, -\frac{1}{3})\)[/tex], not [tex]\((0, 3)\)[/tex].
Answer: False
2. The graph crosses the [tex]\( x \)[/tex]-axis at [tex]\((-2, 0)\)[/tex].
To find where the graph crosses the [tex]\( x \)[/tex]-axis, we need to solve for [tex]\( x \)[/tex] when [tex]\( h(x) = 0 \)[/tex]:
[tex]\[ \frac{2x - 1}{x + 3} = 0. \][/tex]
This equation is zero when the numerator [tex]\(2x - 1 = 0\)[/tex]:
[tex]\[ 2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}. \][/tex]
So, the graph crosses the [tex]\( x \)[/tex]-axis at [tex]\((\frac{1}{2}, 0)\)[/tex], not [tex]\((-2, 0)\)[/tex].
Answer: False
3. The graph has a vertical asymptote at [tex]\( x = -3 \)[/tex].
A vertical asymptote occurs where the denominator is zero and the numerator is not zero. Setting the denominator equal to zero:
[tex]\[ x + 3 = 0 \implies x = -3. \][/tex]
Thus, there is a vertical asymptote at [tex]\( x = -3 \)[/tex].
Answer: True
4. The graph has a hole at [tex]\( (3, 0) \)[/tex].
A hole occurs in the graph when both the numerator and the denominator have a common factor that cancels out, leading to an undefined point. Here, [tex]\(h(x) = \frac{2x - 1}{x + 3}\)[/tex] does not have such a factor that cancels out. Specifically, substituting [tex]\( x = 3 \)[/tex] into [tex]\( h(x) \)[/tex]:
[tex]\[ h(3) = \frac{2(3) - 1}{3 + 3} = \frac{6 - 1}{6} = \frac{5}{6} \neq 0. \][/tex]
There is no hole at [tex]\( (3, 0) \)[/tex].
Answer: False
5. The graph has a horizontal asymptote at [tex]\( y = 2 \)[/tex].
To find the horizontal asymptote, we look at the end behavior of the function as [tex]\( x \)[/tex] approaches infinity. For rational functions, the horizontal asymptote is determined by the ratio of the leading coefficients of the numerator and the denominator:
[tex]\[ \lim_{{x \to \infty}} \frac{2x - 1}{x + 3} = \frac{2x}{x} = 2. \][/tex]
So, the horizontal asymptote is [tex]\( y = 2 \)[/tex].
Answer: True
6. The domain of the function does not include the values [tex]\( x = -3 \)[/tex] and [tex]\( x = 3 \)[/tex].
The domain of the function is determined by the values of [tex]\( x \)[/tex] for which the denominator is not zero. The denominator is zero at [tex]\( x = -3 \)[/tex], making [tex]\( x = -3 \)[/tex] not part of the domain. However, there is no issue at [tex]\( x = 3 \)[/tex]:
[tex]\[ x + 3 \neq 0 \implies x \neq -3. \][/tex]
The domain excludes [tex]\( x = -3 \)[/tex], but [tex]\( x = 3 \)[/tex] is included in the domain.
Answer: False
Summarizing:
\begin{tabular}{|l|l|l|}
\hline The graph crosses the [tex]\( y \)[/tex]-axis at [tex]$(0,3)$[/tex]. & true & \textbf{false} \\
\hline The graph crosses the [tex]\( x \)[/tex]-axis at [tex]$(-2,0)$[/tex]. & true & \textbf{false} \\
\hline The graph has a vertical asymptote at [tex]$x=-3$[/tex]. & \textbf{true} & false \\
\hline The graph has a hole at [tex]$(3,0)$[/tex]. & true & \textbf{false} \\
\hline The graph has a horizontal asymptote at [tex]$y=2$[/tex]. & \textbf{true} & false \\
\hline The domain of the function does not include the values [tex]$x=-3$[/tex] and [tex]$x=3$[/tex]. & true & \textbf{false} \\
\hline
\end{tabular}
