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Question 40

Use a table with values [tex]$x=\{-2, -1, 0, 1, 2\}$[/tex] to graph the quadratic function [tex]$y=-2x^2$[/tex].


Sagot :

Sure, let's build a table of values for the quadratic function [tex]\( y = -2x^2 \)[/tex] using the given [tex]\( x \)[/tex] values and then graph the function.

### Step-by-Step Solution

1. Identify the given [tex]\( x \)[/tex] values:
[tex]\[ x = \{ -2, -1, 0, 1, 2 \} \][/tex]

2. Calculate the corresponding [tex]\( y \)[/tex] values using the function [tex]\( y = -2x^2 \)[/tex]:
- For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -2(-2)^2 = -2(4) = -8 \][/tex]
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = -2(-1)^2 = -2(1) = -2 \][/tex]
- For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = -2(0)^2 = -2(0) = 0 \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = -2(1)^2 = -2(1) = -2 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ y = -2(2)^2 = -2(4) = -8 \][/tex]

3. Construct the table of values:

[tex]\[ \begin{array}{c|c} x & y = -2x^2 \\ \hline -2 & -8 \\ -1 & -2 \\ 0 & 0 \\ 1 & -2 \\ 2 & -8 \\ \end{array} \][/tex]

4. Plot these points on a coordinate system:
- [tex]\((-2, -8)\)[/tex]
- [tex]\((-1, -2)\)[/tex]
- [tex]\((0, 0)\)[/tex]
- [tex]\((1, -2)\)[/tex]
- [tex]\((2, -8)\)[/tex]

5. Draw the quadratic curve:
- Plot the points on the coordinate plane.
- Draw a smooth, continuous curve passing through all the plotted points.

The graph will be a downward-opening parabola since the coefficient of [tex]\( x^2 \)[/tex] is negative.

This table and graph represent the function [tex]\( y = -2x^2 \)[/tex] with the specified [tex]\( x \)[/tex] values.