Question

3
c(x) = x³
g(x) = ¼c(x)
reference points on c(x)
(0, 0)
(1, 1)
(2, 16)
↓
↓
↓
corresponding points on g(x)

topic 3 characteristics of polynomial functions
skills practice continued

Explanation:

Step1: Analyze the function transformation

Given \( g(x)=\frac{1}{4}c(x) \) and \( c(x)=x^3 \), so \( g(x)=\frac{1}{4}x^3 \). We need to find the corresponding points on \( g(x) \) for the given points on \( c(x) \).

Step2: Find the corresponding point for (0,0)

For \( x = 0 \), substitute into \( g(x) \): \( g(0)=\frac{1}{4}\times0^3 = 0 \). So the corresponding point is (0, 0).

Step3: Find the corresponding point for (1,1)

For \( x = 1 \), substitute into \( g(x) \): \( g(1)=\frac{1}{4}\times1^3=\frac{1}{4} \)? Wait, no, wait the table in the problem has (1,1) on \( c(x) \) and we need to check the graph. Wait, maybe I misread. Wait the function is \( g(x)=\frac{1}{4}c(x) \), so if \( c(x) \) has a point \( (x,y) \), then \( g(x) \) has \( (x, \frac{1}{4}y) \)? Wait no, wait the original \( c(x)=x^3 \), so \( c(1)=1^3 = 1 \), \( g(1)=\frac{1}{4}\times1 = \frac{1}{4} \)? But the graph shows \( c(x) \) with (1,1) and \( g(x) \) with (1, something). Wait maybe the function is \( g(x) = 4c(x) \)? Wait the problem says \( g(x)=\frac{1}{4}c(x) \)? Wait no, looking at the graph, the black line has points (-2,-8), (-1,-1), (0,0), (1,1), (2,8) which is \( c(x)=x^3 \). The other line (maybe \( g(x) \))? Wait the table has reference points on \( c(x) \): (0,0), (1,1), (2,16)? Wait no, the table is:

Reference Points on \( c(x) \): (0,0), (1,1), (2,16)? Wait no, the user's table:

| Reference Points on \( c(x) \) | (0,0) | (1,1) | (2,16) |
| --- | --- | --- | --- |
| Corresponding Points on \( g(x) \) |? |? |? |

Wait maybe I misread the function. Wait the problem says \( c(x)=x^3 \), \( g(x)=\frac{1}{4}c(x) \)? No, if \( c(2)=8 \) (since \( c(x)=x^3 \), \( c(2)=8 \)), but the table has (2,16) on \( c(x) \). Wait maybe \( c(x) \) is not \( x^3 \) but another function. Wait the graph: the black curve passes through (-2,-8), (-1,-1), (0,0), (1,1), (2,8) which is \( y = x^3 \). The other line (maybe \( g(x) \)) passes through (-2,-2), (-1,-0.25), (0,0), (1,0.25), (2,2)? No, the grid: each square is 1 unit. Wait the table has (2,16) on \( c(x) \), so \( c(2)=16 \), so \( c(x) \) is not \( x^3 \), maybe \( c(x)=x^4 \)? No, \( x^4 \) is even. Wait maybe the function is \( g(x) = 4c(x) \). Let's check: if \( c(2)=4 \), then \( g(2)=16 \). Wait the table has (2,16) on \( c(x) \)? No, the table is "Reference Points on \( c(x) \)" with (0,0), (1,1), (2,16), and we need to find corresponding on \( g(x) \) where \( g(x)=\frac{1}{4}c(x) \)? No, if \( g(x)=\frac{1}{4}c(x) \), then for (2,16) on \( c(x) \), \( g(2)=\frac{1}{4}\times16 = 4 \). For (1,1) on \( c(x) \), \( g(1)=\frac{1}{4}\times1 = 0.25 \). For (0,0), \( g(0)=0 \). But the graph shows two lines: one is \( c(x) \) (passing through (1,1), (2,8)) and the other (maybe \( g(x) \)) passing through (1,4), (2,32)? No, I think I made a mistake. Wait the problem says \( g(x)=\frac{1}{4}c(x) \) or \( g(x)=4c(x) \)? Let's re-express:

Looking at the table, reference points on \( c(x) \): (0,0), (1,1), (2,16). Corresponding points on \( g(x) \):

For (0,0): \( g(0)=\frac{1}{4}\times0 = 0 \), so (0,0).

For (1,1): \( g(1)=\frac{1}{4}\times1 = 0.25 \), but the graph has a line with (1,4)? Wait no, the black line is \( c(x) \) with (1,1), (2,8), and the other line (maybe \( g(x) \)) has (1,4), (2,32)? No, the grid: each square is 1 unit. Wait the problem's graph: the x-axis and y-axis are grid lines. The black curve (c(x)): (-2,-8), (-1,-1), (0,0), (1,1), (2,8) (since (-2)^3=-8, (-1)^3=-1, 1^3=1, 2^3=8). The other line (g(x)): let's see the points. The line with (-2,-2), (-1,-0.25), (0,0), (1,0.25), (2,2) would be \( y=\frac{1}{4}x^3 \), but the table has (2,16) on c(x). Wait maybe the function is \( g(x)=4c(x) \). Then for (2,4) on c(x) (since c(2)=8, no). Wait I think there's a misprint, but following the function \( g(x)=\frac{1}{4}c(x) \), let's compute the corresponding points:

- For (0,0) on c(x): \( g(0)=\frac{1}{4}\times0 = 0 \), so (0,0).
- For (1,1) on c(x): \( g(1)=\frac{1}{4}\times1 = \frac{1}{4} \), but the table in the problem might have a typo, or maybe the function is \( g(x)=4c(x) \). Wait the point (2,16) on c(x): if \( g(x)=4c(x) \), then \( g(2)=4\times4=16 \)? No, c(2)=8, 4*8=32. Wait I'm confused. Wait the graph shows c(x) as \( y=x^3 \) (points (-2,-8), (-1,-1), (0,0), (1,1), (2,8)) and g(x) as another line. Let's check the table:

Reference Points on c(x): (0,0), (1,1), (2,16)

Corresponding Points on g(x):

- (0,0): since \( g(0)=\frac{1}{4}\times0 = 0 \), so (0,0).
- (1,1): \( g(1)=\frac{1}{4}\times1 = 0.25 \), but maybe the table has (1,1) on c(x) and we need to find g(1). Wait the graph's g(x) (the other line) has points (-2,-2), (-1,-0.25), (0,0), (1,0.25), (2,2), which is \( y=\frac{1}{4}x^3 \), so for x=1, y=0.25; x=2, y=2. But the table has (2,16) on c(x), which doesn't match \( x^3 \). So maybe the c(x) in the table is different. Anyway, following the function \( g(x)=\frac{1}{4}c(x) \), the corresponding points are:

- (0,0) on c(x) → (0,0) on g(x)
- (1,1) on c(x) → (1, \(\frac{1}{4}\times1\)) = (1, 0.25)
- (2,16) on c(x) → (2, \(\frac{1}{4}\times16\)) = (2, 4)

But the graph shows g(x) with (2,8) scaled? Wait no, the black line is c(x)=x^3 (points (-2,-8), (-1,-1), (0,0), (1,1), (2,8)), and the other line (maybe g(x)) has points (-2,-2), (-1,-0.25), (0,0), (1,0.25), (2,2), which is \( y=\frac{1}{4}x^3 \), so the corresponding points are (0,0), (1, 0.25), (2, 2). But the table in the problem has (2,16) on c(x), which is inconsistent with c(x)=x^3. Maybe the function is \( g(x)=4c(x) \), then:

- (0,0) → (0,0)
- (1,1) → (1,4)
- (2,4) → (2,16) (since 4*4=16). Ah, that makes sense! So maybe c(x)=x^2? No, c(x)=x^2 would be even, but the graph is odd. Wait c(x)=x^3, g(x)=4x^3:

- c(2)=8, g(2)=32, no. Wait the table has (2,16) on c(x), so c(2)=16, so c(x)=x^4? No, x^4 at x=2 is 16, yes! c(x)=x^4, then g(x)=\(\frac{1}{4}\)x^4. Then:

- c(0)=0^4=0, g(0)=0
- c(1)=1^4=1, g(1)=\(\frac{1}{4}\)
- c(2)=2^4=16, g(2)=\(\frac{1}{4}\times16 = 4\)

Yes, that matches the table: c(x) has (2,16), so c(x)=x^4, g(x)=\(\frac{1}{4}\)x^4. So the corresponding points are:

- (0,0) on c(x) → (0,0) on g(x)
- (1,1) on c(x) → (1, \(\frac{1}{4}\)) on g(x)
- (2,16) on c(x) → (2,4) on g(x)

But the graph shows a cubic function, which is odd, while x^4 is even. So there's a discrepancy, but following the table:

So the corresponding points are (0,0), (1, 0.25), (2,4). But maybe the problem has a typo, and the function is g(x)=4c(x), then:

- (0,0) → (0,0)
- (1,1) → (1,4)
- (2,4) → (2,16) (but c(x) would be x^2, c(2)=4, g(2)=16). That matches the table: c(x) has (2,4) (but table says (2,16)), no. I think the intended function is g(x)=4c(x), so:

If c(x) has (0,0), (1,1), (2,4), then g(x)=4c(x) has (0,0), (1,4), (2,16). But the graph shows c(x) as x^3, so maybe the table is wrong. Anyway, based on the function \( g(x)=\frac{1}{4}c(x) \), the corresponding points are:

| Reference Points on \( c(x) \) | (0,0) | (1,1) | (2,16) |
| --- | --- | --- | --- |
| Corresponding Points on \( g(x) \) | (0,0) | (1, 0.25) | (2,4) |

But the graph's g(x) (the line with (-2,-2), (-1,-0.25), (0,0), (1,0.25), (2,2)) has (2,2), which is \(\frac{1}{4}\times8\) (since c(2)=8 for x^3), so c(x)=x^3, g(x)=\(\frac{1}{4}\)x^3, so c(2)=8, g(2)=2, but the table has (2,16) on c(x), which is inconsistent. So maybe the table has a typo, and (2,8) on c(x), then g(2)=2.

Assuming the table should have (2,8) on c(x) (since c(x)=x^3), then:

- (0,0) → (0,0)
- (1,1) → (1, 0.25)
- (2,8) → (2, 2)

But the problem's table has (2,16), so I think there's a mistake, but following the function \( g(x)=\frac{1}{4}c(x) \), the corresponding points are (0,0), (1, 0.25), (2,4).

Answer:

The corresponding points on \( g(x) \) are (0, 0), (1, \(\frac{1}{4}\)), and (2, 4). (If we assume the table's (2,16) is a typo and should be (2,8), then (2,2). But based on the given function \( g(x)=\frac{1}{4}c(x) \) and table points, the answers are (0,0), (1, 0.25), (2,4).)