Math  /  Algebra

QuestionComplete the table of values and use to plot the graph with equation y=32xy=-\frac{3}{2} x \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & -2 & -1 & 0 & 1 & 2 & 3 \\ \hlineyy & 3 & \square & & -1.5 & & \\ \hline \end{tabular}

Studdy Solution

STEP 1

What is this asking? We need to fill in the missing *y* values in the table given the equation y=32xy = -\frac{3}{2}x and then plot the points on a graph. Watch out! Don't forget that a negative number multiplied by a negative number results in a positive number!
Also, remember that fractions can be converted to decimals (and vice-versa).

STEP 2

1. Calculate Missing *y* Values
2. Plot the Points

STEP 3

Let's **start** by understanding our equation: y=32xy = -\frac{3}{2} x.
This tells us that the *y*-value is equal to negative three-halves multiplied by the *x*-value.

STEP 4

When x=1x = \mathbf{-1}, let's plug it into our equation: y=32(1)y = -\frac{3}{2} \cdot (-1).
Multiplying a negative by a negative gives us a positive, and multiplying fractions is straightforward, so y=32y = \mathbf{\frac{3}{2}}.
This can also be written as y=1.5y = \mathbf{1.5}.

STEP 5

When x=0x = \mathbf{0}, we have y=320y = -\frac{3}{2} \cdot 0.
Anything multiplied by zero is zero, so y=0y = \mathbf{0}.

STEP 6

When x=2x = \mathbf{2}, we have y=322y = -\frac{3}{2} \cdot 2.
We can think of this as y=322y = -\frac{3 \cdot 2}{2}.
Since we're multiplying and dividing by **2**, we can divide to one, so y=3y = \mathbf{-3}.

STEP 7

When x=3x = \mathbf{3}, we have y=323y = -\frac{3}{2} \cdot 3.
We can think of this as y=332=92y = -\frac{3 \cdot 3}{2} = -\frac{9}{2}, which simplifies to y=4.5y = \mathbf{-4.5}.

STEP 8

Our completed table looks like this: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & -2 & -1 & 0 & 1 & 2 & 3 \\ \hlineyy & 3 & 1.5 & 0 & -1.5 & -3 & -4.5 \\ \hline \end{tabular}

STEP 9

Now, we'll take each (x,y)(x, y) pair from our table and plot it as a point on a graph.
Remember, the *x*-value determines the horizontal position, and the *y*-value determines the vertical position.

STEP 10

Our points are (2,3)(-2, 3), (1,1.5)(-1, 1.5), (0,0)(0, 0), (1,1.5)(1, -1.5), (2,3)(2, -3), and (3,4.5)(3, -4.5).
Plot each of these points, and you'll see they form a straight line!

STEP 11

The completed table is: \begin{tabular}{|c|c|c|c|c|c|c|} \hlinexx & -2 & -1 & 0 & 1 & 2 & 3 \\ \hlineyy & 3 & 1.5 & 0 & -1.5 & -3 & -4.5 \\ \hline \end{tabular} And the graph should be a straight line passing through all the plotted points from this table.

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