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
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.
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.
To complete the table of values for the function y = -(x - 1)^2 + 4, we need to substitute each x value into the function and evaluate the corresponding y value.
Let's do that.
When x = -1:
y = -((-1) - 1)^2 + 4
= -(-2)^2 + 4
= -(4) + 4
= 0
So, when x = -1, y = 0.
When x = 0:
y = -(0 - 1)^2 + 4
= -(-1)^2 + 4
= -(1) + 4
= 3
When x = 1:
y = -(1 - 1)^2 + 4
= -0^2 + 4
= 0 + 4
= 4
When x = 2:
y = -(2 - 1)^2 + 4
= -(1)^2 + 4
= -1 + 4
= 3
When x = 3:
y = -(3 - 1)^2 + 4
= -(2)^2 + 4
= -4 + 4
= 0
Now we have completed the table:
x | y
--------------
-1 | 0
0 | 3
1 | 4
2 | 3
3 | 0
Next, we will plot these values on a graph to analyze the linearity of the function.