Question

7 which table shows a function? a x y 1 -2 1 0 1 1 1 3 b x y -3 -12 -1 -8 1 -4 3 0 c x y 1 7 2 5 3 3 4 7 d x y -3 0 -3 0 0 3 0 -3 8 which equation defines a linear function? a $y = \frac{2}{4}x + 12$ b $y = x^2 + 4x - 6$ c $x^2 + y^2 = 16$ d $\frac{1}{x^2} + \frac{1}{y^2} = 4$ 9 between which interval does the gra increase, remain constant, and then decrease? graph of a function on a coordinate plane a $2 \leq x \leq 5$ b $2 \leq x \leq -2$ c $1 \leq x \leq 6$ d $-2 \leq x \leq 2$

Explanation:

Question 7

Step1: Recall function definition

A function is a relation where each input (x - value) has exactly one output (y - value).

Step2: Analyze Option A

In table A, \(x = 1\) has \(y=-2,0,1,3\). So one \(x\) has multiple \(y\) - values. Not a function.

Step3: Analyze Option B

In table B, \(x=-3\) gives \(y = - 12\), \(x=-1\) gives \(y=-8\), \(x = 1\) gives \(y=-4\), \(x = 3\) gives \(y = 0\). Each \(x\) has one \(y\) - value.

Step4: Analyze Option C

In table C, \(x = 1\) gives \(y = 7\), \(x = 2\) gives \(y = 5\), \(x = 3\) gives \(y = 3\), \(x = 4\) gives \(y = 7\). Even though \(y = 7\) repeats for different \(x\) ( \(x = 1\) and \(x = 4\) ), each \(x\) has one \(y\) - value. Wait, but let's check again. Wait, the definition is each \(x\) has one \(y\), multiple \(x\) can have the same \(y\). But wait, let's check Option D as well.

Step5: Analyze Option D

In table D, \(x=-3\) has \(y = 0\) (twice, but same \(y\), but \(x = 0\) has \(y = 3\) and \(y=-3\). So \(x = 0\) has two \(y\) - values. Not a function.

Wait, there is a mistake in the initial analysis of Option C. Wait, in Option B, all \(x\) values are unique and each has one \(y\). In Option C, \(x = 1,2,3,4\) are all unique and each has one \(y\). But wait, the problem is that in Option B, the \(x\) values are \(-3,-1,1,3\) (all distinct) and each has a unique \(y\). In Option C, \(x = 1,2,3,4\) (distinct) and each has a unique \(y\) (except \(y = 7\) for \(x = 1\) and \(x = 4\), but that's allowed as long as each \(x\) has one \(y\)). But wait, the original tables:

Wait, Option A: \(x = 1\) repeated with different \(y\) - not function.

Option B: \(x=-3,y=-12\); \(x=-1,y=-8\); \(x = 1,y=-4\); \(x = 3,y = 0\). Each \(x\) is unique, each has one \(y\) - function.

Option C: \(x = 1,y = 7\); \(x = 2,y = 5\); \(x = 3,y = 3\); \(x = 4,y = 7\). Each \(x\) has one \(y\) - function.

Option D: \(x=-3\) (twice, but \(y = 0\) both times, but \(x = 0\) has \(y = 3\) and \(y=-3\) - not function.

Wait, this is a problem. Wait, maybe I misread the tables. Let me re - check:

Option B table:

\(x\): \(-3\), \(y\): \(-12\)

\(x\): \(-1\), \(y\): \(-8\)

\(x\): \(1\), \(y\): \(-4\)

\(x\): \(3\), \(y\): \(0\)

All \(x\) values are distinct, each has one \(y\) - function.

Option C table:

\(x\): \(1\), \(y\): \(7\)

\(x\): \(2\), \(y\): \(5\)

\(x\): \(3\), \(y\): \(3\)

\(x\): \(4\), \(y\): \(7\)

All \(x\) values are distinct, each has one \(y\) - function. But this can't be. Wait, maybe there is a typo. Wait, the original problem:

Wait, the user's image:

Option B: \(x\) values: \(-3,-1,1,3\); \(y\) values: \(-12,-8,-4,0\)

Option C: \(x\) values: \(1,2,3,4\); \(y\) values: \(7,5,3,7\)

Ah, in Option C, \(x = 1\) and \(x = 4\) have the same \(y\) - value, but that's allowed. The key is each \(x\) has one \(y\). So both B and C seem to be functions? But that's not possible. Wait, maybe I made a mistake. Wait, the definition of a function is that for every \(x\) in the domain, there is exactly one \(y\) in the range. So in Option B, the domain is \(\{-3,-1,1,3\}\), each \(x\) has one \(y\). In Option C, the domain is \(\{1,2,3,4\}\), each \(x\) has one \(y\). But the options are A, B, C, D. So maybe there is a mistake in my analysis. Wait, let's check the original problem again.

Wait, Option D: \(x=-3\) (two times, but \(y = 0\) both times), \(x = 0\) (two times, \(y = 3\) and \(y=-3\)). So \(x = 0\) has two \(y\) - values - not a function.

Option A: \(x = 1\) has four \(y\) - values - not a function.

So between B and C, both are functions? But that's not possible. Wait…

Step1: Recall linear function form

A linear function is of the form \(y=mx + b\), where \(m\) and \(b\) are constants and the highest power of \(x\) is 1.

Step2: Analyze Option A

\(y=\frac{2}{4}x + 12=\frac{1}{2}x+12\), which is in the form \(y = mx + b\) with \(m=\frac{1}{2}\) and \(b = 12\). Linear function.

Step3: Analyze Option B

\(y=x^{2}+4x - 6\) is a quadratic function (highest power of \(x\) is 2). Not linear.

Step4: Analyze Option C

\(x^{2}+y^{2}=16\) is the equation of a circle. It is not a function (fails vertical line test) and not linear.

Step5: Analyze Option D

\(\frac{1}{x^{2}}+\frac{1}{y^{2}} = 4\) is a rational function and not linear.

Question 9

Step1: Analyze the graph

We need to find the interval where the graph first increases, then remains constant, then decreases.

Step2: Analyze Option A (\(2\leq x\leq5\))

From \(x = 2\) to \(x = 5\): Let's see the graph. At \(x = 2\) to \(x\) around 3, the graph decreases, then remains constant (from \(x = 3\) to \(x = 5\)? No, the graph from \(x = 2\) to \(x = 5\): first decreases, then constant, then decreases? No, the correct interval:

Wait, looking at the graph, the interval \(-2\leq x\leq2\): Let's check the graph. From \(x=-2\) to \(x = 0\) (or some point), the graph increases, then from \(x = 0\) to \(x = 2\) (or some point) remains constant, then decreases? Wait, no. Wait, the option D is \(-2\leq x\leq2\). Wait, let's re - examine.

Wait, the graph: Let's see the key points. The interval where the graph increases, then constant, then decreases. Looking at the graph, the interval \(-2\leq x\leq2\): from \(x=-2\) to \(x = 0\) (increase), then from \(x = 0\) to \(x = 2\) (constant), then from \(x = 2\) onwards (decrease). Wait, no, the correct option is D? Wait, no, let's check the options again.

Wait, the options:

A: \(2\leq x\leq5\)

B: \(2\leq x\leq - 2\) (invalid interval, since \(2\leq x\leq - 2\) is empty)

C: \(1\leq x\leq6\)

D: \(-2\leq x\leq2\)

Wait, the correct interval should be \(-2\leq x\leq2\) (Option D) because in this interval, the graph first increases, then remains constant, then decreases.

Answer:

Question 7: B. Option B (or C, but likely B)
Question 8: A. \(y=\frac{2}{4}x + 12\)
Question 9: D. \(-2\leq x\leq2\)