Question

practice important - the x and y values in every table must be between - 10 and 10 to fit on your graph. group a sample equation: x = 9 slope ______ complete a table of values for the equation. x y domain ____ range ____ group b sample equation: y = 1 slope ____ complete a table of values for the equation. x y domain ____ range ____ group c/d sample equation: - 4y + 2x = 8 slope ____ y - intercept ____ complete a table of values for the equation. x y group e/f sample equation: y = 1/4(x + 2)^2 - 9 circle vertex ______ maximum or minimum complete a table of values for the equation. x y

Explanation:

Group A ($x = 9$)

Step1: Find slope

A vertical line $x = 9$ has an undefined slope.

Step2: Create table of values

Since $x$ is always 9, we can choose $y$-values in the range $- 10\leq y\leq10$. Let $y=-2, - 1,0,1$. The table is:

$x$	$y$
9	-1
9	0
9	1

Step3: Find domain and range

The domain is $\{9\}$ (since $x$ has only one value), and the range is $[-10,10]$ (all possible $y$ - values in the given interval).

Group B ($y = 1$)

Step1: Find slope

A horizontal line $y = 1$ has a slope of 0.

Step2: Create table of values

Since $y$ is always 1, we can choose $x$-values in the range $-10\leq x\leq10$. Let $x=-2,-1,0,1$. The table is:

$x$	$y$
-1	1
0	1
1	1

Step3: Find domain and range

The domain is $[-10,10]$ (all possible $x$ - values in the given interval), and the range is $\{1\}$ (since $y$ has only one value).

Group C/D ($-4y + 2x=8$)

Step1: Rewrite in slope - intercept form

Solve for $y$: $-4y=8 - 2x$, so $y=\frac{1}{2}x - 2$. The slope is $\frac{1}{2}$ and the $y$-intercept is - 2.

Step2: Create table of values

Choose $x$-values in the range $-10\leq x\leq10$. Let $x=-2,0,2,4$.
When $x=-2$, $y=\frac{1}{2}(-2)-2=-1 - 2=-3$.
When $x = 0$, $y=-2$.
When $x = 2$, $y=\frac{1}{2}(2)-2=1 - 2=-1$.
When $x = 4$, $y=\frac{1}{2}(4)-2=2 - 2=0$.
The table is:

$x$	$y$
0	-2
2	-1
4	0

Group E/F ($y=\frac{1}{4}(x + 2)^2-9$)

Step1: Find vertex

For a parabola in the form $y=a(x - h)^2+k$, the vertex is $(h,k)$. Here $h=-2$ and $k=-9$, so the vertex is $(-2,-9)$.
Since $a=\frac{1}{4}>0$, the parabola opens up and has a minimum.

Step2: Create table of values

Choose $x$-values in the range $-10\leq x\leq10$. Let $x=-6,-4,-2,0,2$.
When $x=-6$, $y=\frac{1}{4}(-6 + 2)^2-9=\frac{1}{4}(16)-9=4 - 9=-5$.
When $x=-4$, $y=\frac{1}{4}(-4 + 2)^2-9=\frac{1}{4}(4)-9=1 - 9=-8$.
When $x=-2$, $y=-9$.
When $x = 0$, $y=\frac{1}{4}(0 + 2)^2-9=\frac{1}{4}(4)-9=1 - 9=-8$.
When $x = 2$, $y=\frac{1}{4}(2 + 2)^2-9=\frac{1}{4}(16)-9=4 - 9=-5$.
The table is:

$x$	$y$
-4	-8
-2	-9
0	-8
2	-5

Answer:

Group A

Slope: Undefined
Table of values:

$x$	$y$
9	-1
9	0
9	1

Domain: $\{9\}$
Range: $[-10,10]$

Group B

Slope: 0
Table of values:

$x$	$y$
-1	1
0	1
1	1

Domain: $[-10,10]$
Range: $\{1\}$

Group C/D

Slope: $\frac{1}{2}$
$y$-intercept: - 2
Table of values:

$x$	$y$
0	-2
2	-1
4	0

Group E/F

Vertex: $(-2,-9)$
It has a minimum.
Table of values:

$x$	$y$
-4	-8
-2	-9
0	-8
2	-5