pizzazz 156 155 for hw Nov 15 1 .pdf - .J r CD According to Some ... - Free Printable

Let’s solve each problem step by step.

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Problem 1: Direct Variation

We are told that y varies directly with x. That means:

> y = kx, where k is the constant of variation.

We’re given a table:

| x | y |
|---|---|
| 2 | -4 |
| ? | -6 |
| ? | 8 |

Step 1: Find k using the first pair (x=2, y=-4)

→ y = kx
→ -4 = k * 2
→ k = -4 ÷ 2 = -2

So the equation is: y = -2x

Step 2: Use this to find missing x values.

When y = -6:
→ -6 = -2x
→ x = (-6) ÷ (-2) = 3

When y = 8:
→ 8 = -2x
→ x = 8 ÷ (-2) = -4

✔ So the completed table is:

| x | y |
|----|----|
| 2 | -4 |
| 3 | -6 |
| -4 | 8 |

Equation: y = -2x

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Problem 2: Inverse Variation

We are told that y varies inversely with x. That means:

> y = k/x, or equivalently, xy = k

Given table:

| x | y |
|-----|------|
| 0.5 | -24 |
| ? | -12 |
| ? | 6 |

Step 1: Find k using first pair (x=0.5, y=-24)

→ k = x * y = 0.5 * (-24) = -12

So the equation is: y = -12 / x

Step 2: Find missing x values.

When y = -12:
→ -12 = -12 / x
Multiply both sides by x:
→ -12x = -12
→ x = (-12)/(-12) = 1

When y = 6:
→ 6 = -12 / x
Multiply both sides by x:
→ 6x = -12
→ x = -12 ÷ 6 = -2

✔ Completed table:

| x | y |
|-----|------|
| 0.5 | -24 |
| 1 | -12 |
| -2 | 6 |

Equation: y = -12/x

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Problem 3: Graphs and Asymptotes

We are given two graphs labeled A and B, and we need to identify domain, range, and asymptotes.

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Graph A:

Looking at the graph (even though it's hand-drawn), it appears to be a hyperbola with vertical asymptote at x = 0 and horizontal asymptote at y = 0 — typical for y = 1/x type functions.

But wait — looking at the handwritten notes under Graph A:

It says:
- Domain: {x | x ≠ 0}
- Range: {y | y ≠ 0}
- VA: x = 0
- HA: y = 0

That matches the standard reciprocal function.

Also, there’s a note saying “(x+1)(x-1)” which might suggest factoring, but since the graph looks like y = 1/x shifted? Wait — actually, looking again, the graph seems centered at origin, so likely just y = 1/x.

BUT — in the student’s work, they wrote:

> f(x) = 1/(x² - 1) → which would have VA at x = ±1

Wait — let’s check the graph carefully.

Actually, from the sketch, Graph A has branches in quadrants I and III, crossing near (1,1) and (-1,-1)? No — actually, it looks more like it has vertical asymptotes at x = -1 and x = 1? Because the curve approaches those lines.

Looking at the student’s writing under Graph A:

They wrote:
- Domain: {x | x ≠ -1, 1}
- Range: {y | y ≠ 0} ??? Wait, no — they wrote “Range: {y | y ≥ 0}” — that doesn’t match.

Wait — let me re-express based on what’s drawn.

Actually, looking at the graph labeled “A”, it shows a curve that goes up to infinity as x approaches -1 from left, down to negative infinity as x approaches -1 from right, then comes back up from negative infinity as x approaches 1 from left, and goes to positive infinity as x approaches 1 from right. And it crosses the y-axis at (0, -1).

This suggests the function is something like:

> f(x) = 1 / [(x + 1)(x - 1)] = 1/(x² - 1)

Yes! That makes sense.

So:

Vertical Asymptotes (VA): where denominator = 0 → x² - 1 = 0 → x = ±1

Horizontal Asymptote (HA): degree of numerator (0) < degree of denominator (2) → HA at y = 0

Domain: all real numbers except x = -1 and x = 1 → {x | x ≠ -1, 1}

Range: Since the function can take any value except between some bounds? Let’s think.

For f(x) = 1/(x² - 1):

As x → ±∞, f(x) → 0

At x=0, f(0) = 1/(0 - 1) = -1

The maximum value? Actually, since x² - 1 ≥ -1, but when |x| < 1, x² - 1 is negative, so f(x) is negative; when |x| > 1, f(x) is positive.

Minimum value in positive part: as x→±1+, f(x)→+∞; as x→±∞, f(x)→0+. So positive outputs go from 0+ to ∞.

Negative part: when |x|<1, x² -1 ranges from -1 (at x=0) to 0- (as x→±1-). So f(x) = 1/(negative number between -1 and 0) → so f(x) ≤ -1.

Thus, range is: y ≤ -1 or y > 0 → {y | y ≤ -1 or y > 0}

But in the student’s note, they wrote “Range: {y | y ≥ 0}” — that’s incorrect.

Wait — perhaps I misread the graph.

Alternatively, maybe Graph A is y = 1/x²? But that would be always positive.

Looking again — the graph passes through (0, -1), so it must be negative at x=0.

So yes, f(x) = 1/(x² - 1) fits.

So correct answers for Graph A:

- Domain: all real x except x = -1, 1 → {x | x ≠ -1, 1}
- Range: y ≤ -1 or y > 0 → {y | y ≤ -1 or y > 0}
- VA: x = -1, x = 1
- HA: y = 0

But the student wrote different things — perhaps they made a mistake.

Now look at Graph B.

Graph B: It looks like a hyperbola shifted. The center seems to be at (2, 3)? Or maybe (2, -3)?

Student wrote under Graph B:

- Domain: {x | x ≠ 2}
- Range: {y | y ≠ 3}
- VA: x = 2
- HA: y = 3

And they wrote an equation: f(x) = 3 + 1/(x - 2)

That makes sense!

Because if you have f(x) = a + b/(x - h), then VA at x=h, HA at y=a.

Here, f(x) = 3 + 1/(x - 2)

So:

- VA: x = 2
- HA: y = 3
- Domain: x ≠ 2
- Range: y ≠ 3 (since 1/(x-2) never zero, so f(x) never equals 3)

Perfect.

So for Graph B, student is correct.

For Graph A, student seems to have confused it.

But let’s stick to what’s drawn.

Since the problem says “Identify the domain, range, and asymptotes”, and provides graphs, we should base our answer on the visual.

Assuming Graph A is f(x) = 1/(x² - 1), then:

Final for Graph A:

- Domain: x ≠ -1, 1
- Range: y ≤ -1 or y > 0
- VA: x = -1, x = 1
- HA: y = 0

But to match common textbook problems, sometimes they simplify.

Wait — another possibility: maybe Graph A is y = 1/x, but shifted? No, because it has two vertical asymptotes.

I think my analysis is correct.

However, in the student’s work, for Graph A, they wrote:

“Domain: {x | x ≠ 0}” — wrong

“Range: {y | y ≥ 0}” — wrong

“VA: x=0” — wrong

“HA: y=0” — correct

So they probably misidentified the graph.

But since we’re solving accurately, we’ll go with the correct math.

Perhaps the graph is meant to be y = 1/x, but drawn poorly? Unlikely, because it clearly has two breaks.

Another idea: maybe it’s y = 1/(x(x-1)) or something.

To avoid overcomplicating, and since the student’s own work for Graph B is correct, and for Graph A they have errors, I’ll provide the correct version based on standard interpretation.

But let’s look at the very bottom — there’s a calculation:

“f(x) = 1/(x² - 1)”

And then “L = 1”, “R = -1”, etc. — probably limits.

And “x = -2 or 1” — not sure.

Perhaps for Graph A, it’s intended to be f(x) = 1/(x² - 1), so I’ll go with that.

So summary:

Graph A:
- Equation: f(x) = 1/(x² - 1)
- Domain: {x | x ≠ -1, 1}
- Range: {y | y ≤ -1 or y > 0}
- Vertical Asymptotes: x = -1, x = 1
- Horizontal Asymptote: y = 0

Graph B:
- Equation: f(x) = 3 + 1/(x - 2)
- Domain: {x | x ≠ 2}
- Range: {y | y ≠ 3}
- Vertical Asymptote: x = 2
- Horizontal Asymptote: y = 3

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Problem 4: Rational Inequality

Solve: (x + 2)/(x - 3) ≥ 0

Step 1: Find critical points — where numerator or denominator is zero.

Numerator: x + 2 = 0 → x = -2

Denominator: x - 3 = 0 → x = 3 (excluded from domain)

Step 2: These points divide the number line into intervals:

Interval 1: x < -2

Interval 2: -2 < x < 3

Interval 3: x > 3

Step 3: Test sign in each interval.

Pick test point in Interval 1: x = -3

→ (-3 + 2)/(-3 - 3) = (-1)/(-6) = positive → satisfies ≥ 0

Interval 2: x = 0

→ (0 + 2)/(0 - 3) = 2/(-3) = negative → does not satisfy

Interval 3: x = 4

→ (4 + 2)/(4 - 3) = 6/1 = positive → satisfies

Step 4: Check endpoints.

At x = -2: numerator is 0, so fraction is 0 → 0 ≥ 0 is true → include x = -2

At x = 3: undefined → exclude

So solution: x ≤ -2 or x > 3

In interval notation: (-∞, -2] ∪ (3, ∞)

Student wrote: “x ≤ -2 or x > 3” — correct.

They also wrote “x = -2 or 1” — probably a scratch.

And boxed “x ≤ -2 or x > 3” — good.

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Now, compiling final answers.

──────────────────────────────────────

Final Answer:

Problem 1 (Direct Variation):
Completed table:
x = 3 when y = -6; x = -4 when y = 8
Equation: y = -2x

Problem 2 (Inverse Variation):
Completed table:
x = 1 when y = -12; x = -2 when y = 6
Equation: y = -12/x

Problem 3 (Graphs):
*Graph A:*
Domain: {x | x ≠ -1, 1}
Range: {y | y ≤ -1 or y > 0}
Vertical Asymptotes: x = -1, x = 1
Horizontal Asymptote: y = 0

*Graph B:*
Domain: {x | x ≠ 2}
Range: {y | y ≠ 3}
Vertical Asymptote: x = 2
Horizontal Asymptote: y = 3

Problem 4 (Inequality):
Solution: x ≤ -2 or x > 3
Interval notation: (-∞, -2] ∪ (3, ∞)