Question

the graph of the parent function $f(x) = x^3$ is translated such that the resulting graph can be represented by the function $g(x) = (x - 1)^3 + 1$. which is the graph of $g(x)$?
four graphs are shown here, each with a coordinate grid and a cubic - like curve. the first graph has a curve passing through the origin, the second has a curve crossing the y - axis at (0,2), the third has a curve crossing the x - axis at (2,0), and the fourth has a curve crossing the origin. each graph has x - axis from - 4 to 4 and y - axis from - 4 to 4, with grid lines.

Explanation:

Step1: Recall transformation rules

For a function \( y = f(x - h)+k \), the graph of \( f(x) \) is shifted \( h \) units horizontally (right if \( h>0 \), left if \( h < 0 \)) and \( k \) units vertically (up if \( k>0 \), down if \( k < 0 \)). For \( f(x)=x^{3} \) and \( g(x)=(x - 1)^{3}+1 \), \( h = 1 \) and \( k=1 \). So the parent graph \( y = x^{3} \) (which has a point at \( (0,0) \)) is shifted 1 unit right and 1 unit up. The new point should be \( (0 + 1,0+1)=(1,1) \). Also, we can check the shape: the cubic function \( y=(x - 1)^{3}+1 \) has the same shape as \( y = x^{3} \) (since there's no reflection or vertical/horizontal stretch/compression), just translated. Let's analyze the options:
- First option: The point near the origin is around (0,0) shifted? No, let's check the key point. The parent \( y=x^{3} \) passes through (0,0), (1,1), (-1,-1). For \( g(x)=(x - 1)^{3}+1 \), when \( x = 1 \), \( g(1)=1 \); when \( x=0 \), \( g(0)=( - 1)^{3}+1=-1 + 1=0 \); when \( x = 2 \), \( g(2)=(1)^{3}+1=2 \). Wait, maybe better to find the inflection point (the point where the curve changes concavity, which for \( y=x^{3} \) is at (0,0)). For \( g(x)=(x - 1)^{3}+1 \), the inflection point is at \( (1,1) \). Let's check the graphs:
- First graph: The curve has a point around (0,0) - no, inflection at (0,0) shifted? No.
- Second graph: Inflection at (-1,2)? No.
- Third graph: Let's see, when \( x = 1 \), what's the value? Wait, maybe check the x - intercept or y - intercept. Wait, maybe another approach: the parent function \( y=x^{3} \) is increasing, passes through (0,0), (1,1), (-1,-1). The function \( g(x)=(x - 1)^{3}+1 \) can be expanded as \( g(x)=x^{3}-3x^{2}+3x - 1 + 1=x^{3}-3x^{2}+3x=x(x^{2}-3x + 3) \). The discriminant of \( x^{2}-3x + 3 \) is \( 9-12=-3<0 \), so only real root at \( x = 0 \) (since \( g(0)=0 \)). Wait, no: \( g(x)=(x - 1)^{3}+1 \), set \( g(x)=0 \): \( (x - 1)^{3}=-1\implies x - 1=-1\implies x = 0 \). So x - intercept at (0,0). When \( x = 1 \), \( g(1)=1 \); when \( x=2 \), \( g(2)=2 \); when \( x=-1 \), \( g(-1)=(-2)^{3}+1=-8 + 1=-7 \). Now let's look at the graphs:
- First option: The curve passes through (0,0), and when x increases, it goes up. Let's check the inflection point. The inflection point of \( y=(x - 1)^{3}+1 \) is at \( x = 1 \), \( y = 1 \). Wait, maybe I made a mistake earlier. The inflection point of \( y = x^{3} \) is at (0,0). For \( y=(x - h)^{3}+k \), the inflection point is at (h,k). So for \( g(x)=(x - 1)^{3}+1 \), inflection point is (1,1). Let's check the graphs:
- First graph: The curve has a "bend" around (0,0) - no.
- Second graph: Bend around (-1,2) - no.
- Third graph: Let's see, when x = 1, the y - value: looking at the third graph, when x = 1, y is around 1? Wait, maybe the first graph: wait, no, let's re - evaluate the transformation. The parent function \( f(x)=x^{3} \) is shifted 1 unit to the right (because of \( x-1 \)) and 1 unit up (because of +1). So the point (0,0) on \( f(x) \) moves to (1,1) on \( g(x) \). The point (1,1) on \( f(x) \) moves to (2,2) on \( g(x) \), and (-1,-1) on \( f(x) \) moves to (0,0) on \( g(x) \). Ah! So when x = 0, \( g(0)=(-1)^{3}+1 = 0 \), so (0,0) is on \( g(x) \). When x = 1, \( g(1)=1 \), so (1,1) is on \( g(x) \). Now let's look at the graphs:
- First graph: The curve passes through (0,0), and as x increases, it goes up, with the bend (inflection) around (0,0)? No, wait the inflection point for \( g(x) \) is (1,1). Wait, maybe I confused the inflection point. The derivative of \( g(x) \) is \( g'(x)=3(x - 1)^{2} \), second derivative \( g''(x)=6(x - 1) \). The second derivative is zero at x = 1, so inflection point at (1,1). So the graph should have the inflection point at (1,1). Let's check the first option: the curve has a bend around (0,0) - no. Second option: bend around (-1,2) - no. Third option: let's see, when x = 1, the y - coordinate: looking at the third graph, when x = 1, y is around 1? Wait, maybe the first graph is incorrect. Wait, maybe the key is that the function \( g(x)=(x - 1)^{3}+1 \) can be written as \( g(x)=x^{3}-3x^{2}+3x \). Let's find the y - intercept: when x = 0, y = 0. So y - intercept at (0,0). The x - intercept is also at (0,0) (since \( g(0)=0 \)). Now, let's check the behavior: as \( x\to\infty \), \( g(x)\to\infty \); as \( x\to-\infty \), \( g(x)\to-\infty \), same as \( f(x) \). Now, let's look at the four graphs:
- First graph: The curve passes through (0,0), and when x is 1, y is 1? Wait, the first graph has a curve that starts from the bottom left, comes up through (0,0), and then rises. The inflection point (where the concavity changes) for \( g(x) \) is at x = 1. So before x = 1, the graph is concave down, and after x = 1, concave up? Wait, the second derivative \( g''(x)=6(x - 1) \). So when x < 1, \( g''(x)<0 \) (concave down); when x > 1, \( g''(x)>0 \) (concave up). So the graph should be concave down before x = 1 and concave up after x = 1, with inflection at (1,1). Let's check the first graph: does it have concave down before x = 1 and concave up after? Let's see, at x = 0, it's at (0,0), at x = 1, y = 1, at x = 2, y = 2. The shape: from x < 1 (say x = 0.5), \( g(0.5)=( - 0.5)^{3}+1=-0.125 + 1 = 0.875 \), so the slope at x = 0.5: \( g'(0.5)=3(0.5 - 1)^{2}=3(0.25)=0.75 \). At x = 1.5, \( g(1.5)=(0.5)^{3}+1=0.125 + 1 = 1.125 \), \( g'(1.5)=3(0.5)^{2}=0.75 \). Wait, the first derivative is always non - negative (since \( (x - 1)^{2}\geq0 \)), so the function is always increasing, which matches all cubic graphs (since the coefficient of \( x^{3} \) is positive). Now, the inflection point at (1,1). Let's check the first graph: the curve has a "bend" around (0,0)? No, the inflection point should be at (1,1). Wait, maybe the first graph is the correct one? Wait, no, let's re - express the transformation. The parent function \( y=x^{3} \) is shifted 1 unit right (so replace x with x - 1) and 1 unit up (add 1). So the graph of \( y=x^{3} \) (which goes through (0,0), (1,1), (-1,-1)) becomes \( y=(x - 1)^{3}+1 \), which goes through (1,1), (2,2), (0,0). So the point (0,0) is on \( g(x) \), (1,1) is on \( g(x) \), (2,2) is on \( g(x) \), and (-1,-1) on \( f(x) \) becomes (0,0) on \( g(x) \)? Wait, no: when x = 0, \( g(0)=( - 1)^{3}+1=0 \), so (0,0) is on \( g(x) \). When x = 1, \( g(1)=1 \), so (1,1) is on \( g(x) \). When x = 2, \( g(2)=2 \), so (2,2) is on \( g(x) \). So the graph should pass through (0,0), (1,1), (2,2), with the inflection point at (1,1). Looking at the first graph: it passes through (0,0), and as x increases, it goes up, with the curve bending at (1,1)? Wait, the first graph's curve: when x is 0, y is 0; x = 1, y is 1; x = 2, y is 2. The shape is the same as \( y=x^{3} \) but shifted? Wait, maybe the first option is the correct one. Wait, let's check the other options:
- Second option: The graph passes through (0,2) - no, \( g(0)=0 \), so y - intercept is 0, not 2.
- Third option: The graph passes through (2,0) - but \( g(2)=2 \), so no.
- Fourth option: The graph passes through (0,0) but the shape is shifted left? No, because the transformation is right 1 and up 1.

Wait, maybe I made a mistake in the inflection point. Let's plot some points:

For \( g(x)=(x - 1)^{3}+1 \):

- x = 0: \( g(0)=(-1)^{3}+1 = 0 \)
- x = 1: \( g(1)=0 + 1 = 1 \)
- x = 2: \( g(2)=(1)^{3}+1 = 2 \)
- x = -1: \( g(-1)=(-2)^{3}+1=-8 + 1=-7 \)

So the points (0,0), (1,1), (2,2) are on the graph. Now, looking at the first graph: it has a point at (0,0), and as x increases, it goes up, passing through (1,1) and (2,2) - the shape matches. The other graphs:

- Second graph: Passes through (0,2) - \( g(0)=0 \), so no.
- Third graph: Passes through (2,0) - \( g(2)=2 \), so no.
- Fourth graph: Passes through (0,0) but the left - hand side is different.

So the first graph is the graph of \( g(x)=(x - 1)^{3}+1 \).

Answer:

The first graph (the one with the curve passing through (0,0), rising with the inflection point around (1,1)) is the graph of \( g(x)=(x - 1)^{3}+1 \). (Assuming the first option is the one with the circle at the bottom left, the curve through (0,0), and rising to the right.)