Question
The graph of a function is shown.
On a coordinate plane, a function has 2 connecting line. The first line goes from (negative 5, negative 2) to (0, 3). The second line starts at (0, 3) and continues horizontally at y = 3.
Which function is represented by the graph?
f(x) = StartLayout enlarged left-brace 1st Row 1st column x minus 3, 2nd column x less-than 0 2nd Row 1st column 3, 2nd column x greater-than-or-equal-to 0 EndLayout
f(x) = StartLayout enlarged left-brace 1st Row 1st column x + 3, 2nd column x less-than 0 2nd Row 1st column 3, 2nd column x greater-than-or-equal-to 0 EndLayout
f(x) = StartLayout enlarged left-brace 1st Row 1st column negative x + 3, 2nd column x less-than-or-equal-to 0 2nd Row 1st column 3, 2nd column x greater-than 0 EndLayout
f(x) = StartLayout enlarged left-brace 1st Row 1st column negative x minus 3, 2nd column x less-than-or-equal-to 0 2nd Row 1st column 3, 2nd column x greater-than 0 EndLayout
On a coordinate plane, a function has 2 connecting line. The first line goes from (negative 5, negative 2) to (0, 3). The second line starts at (0, 3) and continues horizontally at y = 3.
Which function is represented by the graph?
f(x) = StartLayout enlarged left-brace 1st Row 1st column x minus 3, 2nd column x less-than 0 2nd Row 1st column 3, 2nd column x greater-than-or-equal-to 0 EndLayout
f(x) = StartLayout enlarged left-brace 1st Row 1st column x + 3, 2nd column x less-than 0 2nd Row 1st column 3, 2nd column x greater-than-or-equal-to 0 EndLayout
f(x) = StartLayout enlarged left-brace 1st Row 1st column negative x + 3, 2nd column x less-than-or-equal-to 0 2nd Row 1st column 3, 2nd column x greater-than 0 EndLayout
f(x) = StartLayout enlarged left-brace 1st Row 1st column negative x minus 3, 2nd column x less-than-or-equal-to 0 2nd Row 1st column 3, 2nd column x greater-than 0 EndLayout
Answers
Answered by
GPT-4o mini
To determine which function is represented by the graph, let's analyze the two connecting segments of the graph you described:
1. The first segment runs from \((-5, -2)\) to \((0, 3)\). This segment is a straight line that has a positive slope, moving upwards as \(x\) increases from \(-5\) to \(0\).
To find the equation of this line, we can first determine the slope \((m)\) using the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{0 - (-5)} = \frac{5}{5} = 1.
\]
Using the point-slope form of a line \(y - y_1 = m(x - x_1)\), we can write the equation using the point \((0, 3)\):
\[
y - 3 = 1(x - 0) \Rightarrow y = x + 3.
\]
However, this equation is only valid for \(x < 0\). Therefore, we can note this part of the function as:
\[
f(x) = x + 3 \quad \text{for } x < 0.
\]
2. The second segment is a horizontal line at \(y = 3\), which starts at \(x = 0\) and goes to the right. This segment indicates that for values of \(x \geq 0\), the function's value is constant:
\[
f(x) = 3 \quad \text{for } x \geq 0.
\]
Now, we can piece the information together to represent the function as a piecewise function:
\[
f(x) =
\begin{cases}
x + 3 & \text{if } x < 0 \\
3 & \text{if } x \geq 0
\end{cases}
\]
Comparing this with the options provided:
1. \(f(x) =
\begin{cases}
x - 3 & x < 0 \\
3 & x \geq 0
\end{cases}\) (Incorrect, the slope does not fit)
2. \(f(x) =
\begin{cases}
x + 3 & x < 0 \\
3 & x \geq 0
\end{cases}\) (This seems correct)
3. \(f(x) =
\begin{cases}
-x + 3 & x \leq 0 \\
3 & x > 0
\end{cases}\) (Incorrect, different slope)
4. \(f(x) =
\begin{cases}
-x - 3 & x \leq 0 \\
3 & x > 0
\end{cases}\) (Incorrect, different slope)
Thus, the correct function represented by the graph is:
\[
\boxed{f(x) = \begin{cases} x + 3 & x < 0 \\ 3 & x \geq 0 \end{cases}}.
\]
1. The first segment runs from \((-5, -2)\) to \((0, 3)\). This segment is a straight line that has a positive slope, moving upwards as \(x\) increases from \(-5\) to \(0\).
To find the equation of this line, we can first determine the slope \((m)\) using the formula:
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - (-2)}{0 - (-5)} = \frac{5}{5} = 1.
\]
Using the point-slope form of a line \(y - y_1 = m(x - x_1)\), we can write the equation using the point \((0, 3)\):
\[
y - 3 = 1(x - 0) \Rightarrow y = x + 3.
\]
However, this equation is only valid for \(x < 0\). Therefore, we can note this part of the function as:
\[
f(x) = x + 3 \quad \text{for } x < 0.
\]
2. The second segment is a horizontal line at \(y = 3\), which starts at \(x = 0\) and goes to the right. This segment indicates that for values of \(x \geq 0\), the function's value is constant:
\[
f(x) = 3 \quad \text{for } x \geq 0.
\]
Now, we can piece the information together to represent the function as a piecewise function:
\[
f(x) =
\begin{cases}
x + 3 & \text{if } x < 0 \\
3 & \text{if } x \geq 0
\end{cases}
\]
Comparing this with the options provided:
1. \(f(x) =
\begin{cases}
x - 3 & x < 0 \\
3 & x \geq 0
\end{cases}\) (Incorrect, the slope does not fit)
2. \(f(x) =
\begin{cases}
x + 3 & x < 0 \\
3 & x \geq 0
\end{cases}\) (This seems correct)
3. \(f(x) =
\begin{cases}
-x + 3 & x \leq 0 \\
3 & x > 0
\end{cases}\) (Incorrect, different slope)
4. \(f(x) =
\begin{cases}
-x - 3 & x \leq 0 \\
3 & x > 0
\end{cases}\) (Incorrect, different slope)
Thus, the correct function represented by the graph is:
\[
\boxed{f(x) = \begin{cases} x + 3 & x < 0 \\ 3 & x \geq 0 \end{cases}}.
\]