50155

Question

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Worked Solution

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Variant 0

DifficultyLevel

569

Question

Here is a table of values for x\large x and y\large y.

x\large x 0 0.5 1 1.5 2
y\large y 0 1 4 9 16

Which of these is a correct rule for y\large y in terms of x\large x?

Worked Solution

By trial and error for each given equation:

Consider  y\large y = 4x\large x2^2
0=4×020 = 4 × 0^2  ✓\checkmark
1=4×0.521 = 4 × 0.5^2  ✓\checkmark
4=4×124 = 4 × 1^2  ✓\checkmark

∴ y\therefore\ \large y = 4x2\large x^2  is the correct rule.

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Here is a table of values for $\large x$ and $\large y$.
>>| $\large x$|0|0.5|1|1.5|2| |:-:|:-:|:-:|:-:|:-:|:-:| | $\large y$ | 0|1|4|9|16|

Which of these is a correct rule for $\large y$ in terms of $\large x$?
workedSolution
By trial and error for each given equation:
Consider  $\large y$ = 4$\large x$$^2$
>>| | | ----------------------- | |$0 = 4 × 0^2$  $\checkmark$| |$1 = 4 × 0.5^2$  $\checkmark$| |$4 = 4 × 1^2$  $\checkmark$|
$\therefore\ \large y$ = 4$\large x^2$  is the correct rule.
correctAnswer
$\large y$ = 4$\large x$$^2$

Answers

Is Correct?Answer
x

y\large y = x\large x

x

y\large y = 2x\large x

x

y\large y = 6x\large x

✓

y\large y = 4x\large x2^2

Variant 1

DifficultyLevel

568

Question

Here is a table of values for x\large x and y\large y.

x\large x 0 0.5 1 1.5 2
y\large y 0 2 8 18 32

Which of these is a correct rule for y\large y in terms of x\large x?

Worked Solution

By trial and error for each given equation:

Consider  y\large y = 8x\large x2^2
0=8×020 = 8 × 0^2  ✓\checkmark
2=8×0.522 = 8 × 0.5^2  ✓\checkmark
8=8×128 = 8 × 1^2  ✓\checkmark

∴ y\therefore\ \large y = 8x2\large x^2  is the correct rule.

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Here is a table of values for $\large x$ and $\large y$.
>>| $\large x$|0|0.5|1|1.5|2| |:-:|:-:|:-:|:-:|:-:|:-:| | $\large y$ | 0|2|8|18|32|

Which of these is a correct rule for $\large y$ in terms of $\large x$?
workedSolution
By trial and error for each given equation:
Consider  $\large y$ = 8$\large x$$^2$
>>| | | ----------------------- | |$0 = 8 × 0^2$  $\checkmark$| |$2 = 8 × 0.5^2$  $\checkmark$| |$8 = 8 × 1^2$  $\checkmark$|
$\therefore\ \large y$ = 8$\large x^2$  is the correct rule.
correctAnswer
$\large y$ = 8$\large x$$^2$

Answers

Is Correct?Answer
x

y\large y = x\large x

x

y\large y = 4x\large x

✓

y\large y = 8x\large x2^2

x

y\large y = 16x\large x

Variant 2

DifficultyLevel

536

Question

Here is a table of values for x\large x and y\large y.

x\large x 0 1 2 3 4
y\large y −1-1 1 3 5 7

Which of these is a correct rule for y\large y in terms of x\large x?

Worked Solution

By trial and error for each given equation:

Consider  y\large y = 2x\large x −- 1
−1=2×0−1-1 = 2 × 0 - 1  ✓\checkmark
1=2×1−11 = 2 × 1 - 1  ✓\checkmark
3=2×2−13 = 2 × 2 - 1  ✓\checkmark
5=2×3−15 = 2 × 3 - 1  ✓\checkmark

∴\therefore  y\large y = 2x\large x −- 1  is the correct rule.

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Here is a table of values for $\large x$ and $\large y$.
>>| $\large x$|0|1|2|3|4| |:-:|:-:|:-:|:-:|:-:|:-:| | $\large y$ | $-1$|1|3|5|7|

Which of these is a correct rule for $\large y$ in terms of $\large x$?
workedSolution
By trial and error for each given equation:
Consider  $\large y$ = 2$\large x$ $-$ 1
>>| | | ----------------------- | |$-1 = 2 × 0 - 1$  $\checkmark$| |$1 = 2 × 1 - 1$  $\checkmark$| |$3 = 2 × 2 - 1$  $\checkmark$| |$5 = 2 × 3 - 1$  $\checkmark$|
$\therefore$  $\large y$ = 2$\large x$ $-$ 1  is the correct rule.
correctAnswer
$\large y$ = 2$\large x$ $-$ 1

Answers

Is Correct?Answer
x

y\large y = −x- \large x

x

y\large y = x\large x −- 1

x

y\large y = x\large x2^2 −- 1

✓

y\large y = 2x\large x −- 1

Variant 3

DifficultyLevel

570

Question

Here is a table of values for x\large x and y\large y.

x\large x 0 0.5 1 1.5 2
y\large y 0 0.5 2 4.5 8

Which of these is a correct rule for y\large y in terms of x\large x?

Worked Solution

By trial and error for each given equation:

Consider  y\large y = 2x\large x2^2
0=2×020 = 2 × 0^2  ✓\checkmark
0.5=2×0.520.5 = 2 × 0.5^2  ✓\checkmark
2=2×122 = 2 × 1^2  ✓\checkmark
4.5=2×1.524.5 = 2 × 1.5^2  ✓\checkmark
8=2×228 = 2 × 2^2  ✓\checkmark

∴ y\therefore\ \large y = 2x\large x2^2  is the correct rule.

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Here is a table of values for $\large x$ and $\large y$.
>>| $\large x$|0|0.5|1|1.5|2| |:-:|:-:|:-:|:-:|:-:|:-:| | $\large y$ | 0| 0.5|2|4.5|8|

Which of these is a correct rule for $\large y$ in terms of $\large x$?
workedSolution
By trial and error for each given equation:
Consider  $\large y$ = 2$\large x$$^2$
>>| | | ----------------------- | |$0 = 2 × 0^2$  $\checkmark$| |$0.5 = 2 × 0.5^2$  $\checkmark$| |$2 = 2 × 1^2$  $\checkmark$| |$4.5 = 2 × 1.5^2$  $\checkmark$| |$8 = 2 × 2^2$  $\checkmark$|
$\therefore\ \large y$ = 2$\large x$$^2$  is the correct rule.
correctAnswer
$\large y$ = 2$\large x$$^2$

Answers

Is Correct?Answer
x

y\large y = 2x2\large x

✓

y\large y = 2x\large x2^2

x

y\large y = 4x\large x

x

y\large y = 4x\large x2^2

Variant 4

DifficultyLevel

571

Question

Here is a table of values for x\large x and y\large y.

x\large x 0 0.5 1 1.5 2
y\large y 0 −- 0.5 −- 2 −- 4.5 −- 8

Which of these is a correct rule for y\large y in terms of x\large x?

Worked Solution

By trial and error for each given equation:

Consider  y\large y = −- 2x\large x2^2
0 = −- 2×022 × 0^2  ✓\checkmark
−- 0.5 = −- 2×0.522 × 0.5^2  ✓\checkmark
−- 2 = −- 2×122 × 1^2  ✓\checkmark
−- 4.5 = −- 2×1.522 × 1.5^2  ✓\checkmark
−- 8 = −- 2×222 × 2^2  ✓\checkmark

∴ y\therefore\ \large y = −- 2x\large x2^2  is the correct rule.

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Here is a table of values for $\large x$ and $\large y$.
>>| $\large x$|0|0.5|1|1.5|2| |:-:|:-:|:-:|:-:|:-:|:-:| | $\large y$ | 0|$-$ 0.5|$-$ 2|$-$ 4.5|$-$ 8|

Which of these is a correct rule for $\large y$ in terms of $\large x$?
workedSolution
By trial and error for each given equation:
Consider  $\large y$ = $-$ 2$\large x$$^2$
>>| | | ----------------------- | |0 = $-$ $2 × 0^2$  $\checkmark$| |$-$ 0.5 = $-$ $2 × 0.5^2$  $\checkmark$| |$-$ 2 = $-$ $2 × 1^2$  $\checkmark$| |$-$ 4.5 = $-$ $2 × 1.5^2$  $\checkmark$| |$-$ 8 = $-$ $2 × 2^2$  $\checkmark$|
$\therefore\ \large y$ = $-$ 2$\large x$$^2$  is the correct rule.
correctAnswer
$\large y$ = $-$ 2$\large x$$^2$

Answers

Is Correct?Answer
✓

y\large y = −- 2x\large x2^2

x

y\large y = −- 2x2\large x

x

y\large y = 2x\large x2^2

x

y\large y = 2x2\large x

Variant 5

DifficultyLevel

558

Question

Here is a table of values for x\large x and y\large y.

x\large x 0 1 2 3 4
y\large y 0 1 4 9 16

Which of these is a correct rule for y\large y in terms of x\large x?

Worked Solution

By trial and error for each given equation:

Consider  y\large y = x\large x2^2
0=020 = 0^2  ✓\checkmark
1=121 = 1^2  ✓\checkmark
4=224 = 2^2  ✓\checkmark
9=329 = 3^2  ✓\checkmark
16=4216 = 4^2  ✓\checkmark

∴ y\therefore\ \large y = x\large x2^2  is the correct rule.

Question Type

Multiple Choice (One Answer)

Variables

Variable nameVariable value
question
Here is a table of values for $\large x$ and $\large y$.
>>| $\large x$|0|1|2|3|4| |:-:|:-:|:-:|:-:|:-:|:-:| | $\large y$ | 0| 1|4|9|16|

Which of these is a correct rule for $\large y$ in terms of $\large x$?
workedSolution
By trial and error for each given equation:
Consider  $\large y$ = $\large x$$^2$
>>| | | ----------------------- | |$0 = 0^2$  $\checkmark$| |$1 = 1^2$  $\checkmark$| |$4 = 2^2$  $\checkmark$| |$9 = 3^2$  $\checkmark$| |$16 = 4^2$  $\checkmark$|
$\therefore\ \large y$ = $\large x$$^2$  is the correct rule.
correctAnswer
$\large y$ = $\large x$$^2$

Answers

Is Correct?Answer
x

y\large y = x\large x

✓

y\large y = x\large x2^2

x

y\large y = 2x2\large x

x

y\large y = 2x2\large x2^2

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