Question

the function $g(x)$ is a transformation of the parent function $f(x) = x^2$. the graph of $g(x)$ is shown below.

(graph of a parabola with vertex at (0, -3), opening upwards)

select which of the following transformations describes the transformation of $f(x)$ that results in $g(x)$.

• $f(x)$ was translated down 3 units.
• $f(x)$ was vertically stretched by a factor of $\frac{1}{2}$ and translated down 3 units.
• $f(x)$ was vertically stretched by a factor of $\frac{1}{2}$ and translated up 3 units.
• $f(x)$ was vertically stretched by a factor of 2 and translated down 3 units.
• $f(x)$ was translated up 3 units.
• $f(x)$ was vertically stretched by a factor of 2 and translated up 3 units.

enter the function $g(x)$.

Explanation:

Step1: Analyze vertical translation

The parent function \( f(x) = x^2 \) has its vertex at \( (0,0) \). The graph of \( g(x) \) has its vertex at \( (0, -3) \), so there is a vertical translation down 3 units.

Step2: Analyze vertical stretch

To check the vertical stretch, we can use a point. For \( f(x) = x^2 \), when \( x = 1 \), \( f(1) = 1 \). For \( g(x) \), when \( x = 1 \), from the graph, \( g(1) = -2 \)? Wait, no, looking at the graph, when \( x = 1 \), the \( y \)-value: wait, the vertex is at \( (0, -3) \), and when \( x = 1 \), let's see the grid. Wait, maybe I made a mistake. Wait, the parent function \( f(x)=x^2 \), if we have a vertical stretch by factor \( a \) and translation down 3, then \( g(x)=a x^2 - 3 \). Let's take a point on \( g(x) \), say \( x = 1 \), what's \( y \)? From the graph, when \( x = 1 \), the point is at \( (1, -2) \)? Wait, no, the graph crosses the \( x \)-axis at \( x = -1 \) and \( x = 1 \)? Wait, no, the vertex is at \( (0, -3) \), so when \( y = 0 \), \( 0 = a x^2 - 3 \), so \( a x^2 = 3 \), and when \( x = 1 \), \( a(1)^2 = 3 \)? No, that can't be. Wait, maybe the vertical stretch is by factor \( \frac{1}{2} \)? Wait, no, let's re - examine. Wait, the correct way: the vertex form of a parabola is \( g(x)=a(x - h)^2 + k \), where \( (h,k) \) is the vertex. Here, \( h = 0 \), \( k=-3 \), so \( g(x)=a x^2-3 \). Let's take a point on the graph, for example, when \( x = \sqrt{6} \)? No, maybe a better approach: the parent function \( f(x)=x^2 \), if we translate down 3, it's \( f(x)=x^2 - 3 \). But the graph of \( g(x) \) seems to have a different shape. Wait, no, maybe the initial thought about the stretch was wrong. Wait, the vertex is at \( (0, -3) \), so the translation is down 3. Now, check the width. The parent function \( y = x^2 \) has a "width" such that at \( y = 1 \), \( x=\pm1 \). For \( g(x) \), when \( y=-2 \) (which is 1 unit above the vertex), what's \( x \)? If \( g(x)=\frac{1}{2}x^2-3 \), then when \( y=-2 \), \( -2=\frac{1}{2}x^2 - 3 \), so \( \frac{1}{2}x^2=1 \), \( x^2 = 2 \), \( x=\pm\sqrt{2}\approx\pm1.414 \). But from the graph, the parabola passes through \( (1, -2) \)? Wait, no, the graph as shown: the vertex is at \( (0, -3) \), and it crosses the \( x \)-axis at \( x=-1 \) and \( x = 1 \)? Wait, when \( y = 0 \), \( 0=a x^2-3 \), so \( x^2=\frac{3}{a} \). If \( x = 1 \) is a root, then \( 1=\frac{3}{a} \), so \( a = 3 \)? No, that's not matching. Wait, maybe I misread the graph. Wait, the user - provided graph: the vertex is at \( (0, -3) \), and the parabola opens upwards. Let's take \( x = 1 \), the \( y \)-coordinate: looking at the grid, each square is 1 unit. So from the vertex \( (0, -3) \), moving right 1 unit (x = 1), the \( y \)-value: let's see, the parabola at \( x = 1 \) is at \( y=-2 \)? So \( g(1)=-2 \). Then using \( g(x)=a x^2-3 \), substitute \( x = 1 \), \( y=-2 \): \( -2=a(1)^2-3 \), so \( a=1 \)? But that would mean no stretch. But the option says "translated down 3 units" is an option. Wait, maybe the initial analysis of the stretch was wrong. Wait, the first option is " \( f(x) \) was translated down 3 units", which would be \( g(x)=x^2-3 \). Let's check the roots: \( x^2-3 = 0\Rightarrow x=\pm\sqrt{3}\approx\pm1.732 \), but the graph in the problem shows the parabola crossing the \( x \)-axis at \( x=-1 \) and \( x = 1 \)? Wait, no, the graph as drawn: the vertex is at \( (0, -3) \), and the parabola passes through \( (1, -2) \)? No, maybe the graph is actually \( g(x)=x^2-3 \), but the options have a stretch. Wait, maybe I made a mistake in the graph reading. Wait, the original problem's graph: let's look again. The \( y \)-axis: the vertex is at \( (0, -3) \), and when \( x = 1 \), the \( y \)-value is \( -2 \), so \( g(1)=-2 \). Then \( g(x)=x^2-3 \) would give \( g(1)=1 - 3=-2 \), which matches. So there is no vertical stretch, only a translation down 3 units. Wait, but the selected option in the original problem was the second one, but maybe that's a mistake. Wait, let's re - evaluate.

Wait, the parent function \( f(x)=x^2 \), vertex at \( (0,0) \). The graph of \( g(x) \) has vertex at \( (0, -3) \), so vertical translation down 3. Now, check the shape: the width of the parabola. For \( f(x)=x^2 \), the distance from the vertex to the point where \( y = 1 \) is \( x=\pm1 \). For \( g(x) \), when \( y=-2 \) (which is 1 unit above the vertex, since vertex is at \( y=-3 \)), \( x=\pm1 \). So the distance from the vertex (x = 0) to \( x = 1 \) when \( y=-2 \) is 1, same as \( f(x) \) when \( y = 1 \) (x = 1). So the vertical stretch factor \( a \) is 1, because \( g(x)=1\times x^2-3=x^2 - 3 \). So the correct transformation is \( f(x) \) was translated down 3 units. But maybe I misread the graph. Wait, the original problem's options: the first option is " \( f(x) \) was translated down 3 units". Let's confirm with the vertex: \( f(x)=x^2 \) vertex (0,0), \( g(x) \) vertex (0, -3), so translation down 3. And the shape: since the coefficient of \( x^2 \) is 1 (same as parent), no stretch. So the correct transformation is translation down 3 units.

Now, for the function \( g(x) \): since it's a translation down 3 units from \( f(x)=x^2 \), \( g(x)=x^2-3 \). Wait, but let's check with the graph. If \( g(x)=x^2 - 3 \), then when \( x = 1 \), \( g(1)=1 - 3=-2 \), which matches the graph (the point at \( (1, -2) \)). When \( x = 0 \), \( g(0)=-3 \), which is the vertex. So that's correct.

Step1 (revised): Determine vertical translation

The vertex of \( f(x)=x^2 \) is \( (0,0) \), and the vertex of \( g(x) \) is \( (0, -3) \). So \( f(x) \) is translated down 3 units.

Step2 (revised): Determine vertical stretch

The coefficient of \( x^2 \) in \( g(x) \) is 1 (same as \( f(x) \)) because the "width" of the parabola (distance from vertex to the point with \( y \)-value 1 unit above vertex) is the same as \( f(x) \). So there is no vertical stretch.

Answer:

For the transformation: \( f(x) \) was translated down 3 units.
For the function \( g(x) \): \( g(x)=x^2 - 3 \)