To determine if the given function \( y = -(x - 1)^2 + 4 \) is linear or nonlinear, and then to show why it is linear or nonlinear, we need to first calculate the values of \( y \) for different inputs of \( x \), and then plot them on a graph.
We will start with the given values of \( x \) to complete the table.
For \( x = -1 \):
\[ y = -((-1) - 1)^2 + 4 = -(0)^2 + 4 = -0 + 4 = 4 \]
For \( x = 0 \):
\[ y = -((0) - 1)^2 + 4 = -(1)^2 + 4 = -1 + 4 = 3 \]
For \( x = 1 \):
\[ y = -((1) - 1)^2 + 4 = -(0)^2 + 4 = -0 + 4 = 4 \]
For \( x = 2 \):
\[ y = -((2) - 1)^2 + 4 = -(1)^2 + 4 = -1 + 4 = 3 \]
For \( x = 3 \):
\[ y = -((3) - 1)^2 + 4 = -(2)^2 + 4 = -4 + 4 = 0 \]
Now, the table of values looks like this:
\[
\begin{array}{cc}
x & y \\
\hline
-1 & 4 \\
0 & 3 \\
1 & 4 \\
2 & 3 \\
3 & 0 \\
\end{array}
\]
Plotting these points on a coordinate grid, we can observe the shape of the graph:
1. (-1, 4)
2. (0, 3)
3. (1, 4)
4. (2, 3)
5. (3, 0)
The graph of \( y = -(x - 1)^2 + 4 \) will form a downward-opening parabola with the vertex at \( (1, 4) \).
This function is not linear because it does not create a straight line when graphed, and the rate of change is not constant. In a linear graph, the change in \( y \) is consistent for each unit increase in \( x \). However, in this case, the rate of change in \( y \) varies, indicating that the function is nonlinear. Therefore, \( y = -(x - 1)^2 + 4 \) is a nonlinear function due to its parabolic shape.
Challenge Decide whether the function is linear or nonlinear using both a table and a graph. Use pencil and paper. Find a linear function. Make a table and a graph for the function to explain why it is linear.
yequals
minus(xnegative 1)squaredplus 4
Question content area bottom
Part 1
Complete the table of values for the function.
x
y
negative 1
enter your response here
0
enter your response here
1
enter your response here
2
enter your response here
3
enter your response here
3 answers
no
Apologies for the confusion. I see that I mistakenly indicated we would explain why the function is linear when in fact the function provided is not linear. Let's correct that.
The function given in your question is \( y = -(x - 1)^2 + 4 \). This is a quadratic function, not a linear function, as it includes a squared term.
A linear function has the form \( y = mx + b \) where \( m \) is the slope and \( b \) is the y-intercept. There are no squared terms in a linear function.
To demonstrate that the given function is nonlinear, I will complete the table using the original values for \( x \) again, and you can graph these points on a coordinate grid to see the shape of the function.
\[
\begin{array}{cc}
x & y \\
\hline
-1 & -((-1) - 1)^2 + 4 = -(0)^2 + 4 = 4 \\
0 & -((0) - 1)^2 + 4 = -(1)^2 + 4 = 3 \\
1 & -((1) - 1)^2 + 4 = -(0)^2 + 4 = 4 \\
2 & -((2) - 1)^2 + 4 = -(1)^2 + 4 = 3 \\
3 & -((3) - 1)^2 + 4 = -(2)^2 + 4 = 0\\
\end{array}
\]
The graph of these points will show a parabola, which is the graph of a quadratic function, not a straight line as would be seen with a linear function. Thus, by both the table of values and the graphical representation, we can confirm that the function is indeed nonlinear.
The function given in your question is \( y = -(x - 1)^2 + 4 \). This is a quadratic function, not a linear function, as it includes a squared term.
A linear function has the form \( y = mx + b \) where \( m \) is the slope and \( b \) is the y-intercept. There are no squared terms in a linear function.
To demonstrate that the given function is nonlinear, I will complete the table using the original values for \( x \) again, and you can graph these points on a coordinate grid to see the shape of the function.
\[
\begin{array}{cc}
x & y \\
\hline
-1 & -((-1) - 1)^2 + 4 = -(0)^2 + 4 = 4 \\
0 & -((0) - 1)^2 + 4 = -(1)^2 + 4 = 3 \\
1 & -((1) - 1)^2 + 4 = -(0)^2 + 4 = 4 \\
2 & -((2) - 1)^2 + 4 = -(1)^2 + 4 = 3 \\
3 & -((3) - 1)^2 + 4 = -(2)^2 + 4 = 0\\
\end{array}
\]
The graph of these points will show a parabola, which is the graph of a quadratic function, not a straight line as would be seen with a linear function. Thus, by both the table of values and the graphical representation, we can confirm that the function is indeed nonlinear.
1 answer