Question

use the graph of f(x) = x² to write an equation for the function represented by each graph.
(a)
graph of a parabola opening upwards with vertex at (0, -3) and x-intercepts at -1.5 and 1.5 (approx)
y =
(b)
graph of a parabola opening downwards with vertex at (-1, 4) and x-intercepts at -3 and 1 (approx)
y =

Explanation:

Part (a)

Step 1: Analyze the transformation of \( f(x) = x^2 \)

The parent function is \( f(x) = x^2 \), which has its vertex at \( (0,0) \). The given graph in part (a) has its vertex at \( (0, -3) \) (by observing the graph, the minimum point is at \( y = -3 \) on the y - axis). There is no horizontal shift (since the vertex is on the y - axis) and no reflection (the parabola opens upwards, same as \( f(x)=x^2 \)). The general form of a vertical shift of \( f(x)=x^2 \) is \( g(x)=x^2 + k \), where \( k \) is the vertical shift. If \( k<0 \), the graph shifts down. Here, the vertex is at \( (0,-3) \), so \( k=-3 \).

Step 2: Write the equation

Using the vertical shift formula, the equation of the function is \( y=x^2 - 3 \).

Part (b)

Step 1: Analyze the transformation of \( f(x) = x^2 \)

The parent function \( f(x)=x^2 \) opens upwards. The given graph in part (b) opens downwards, so there is a reflection over the x - axis. The general form for a reflection over the x - axis is \( g(x)=-x^2 \). Now, we need to check the vertical shift or stretch. The vertex of the parent function \( f(x)=x^2 \) is at \( (0,0) \). The vertex of the given graph in part (b) is at \( (0,4) \)? Wait, no, let's re - examine. The graph is a downward - opening parabola. Let's find the vertex. From the graph, the maximum point (vertex) seems to be at \( (0,4) \)? Wait, no, looking at the grid, the vertex is at \( (0,4) \)? Wait, the graph passes through \( (0,3) \)? Wait, no, let's use the standard form of a quadratic function \( y = a(x - h)^2+k \), where \( (h,k) \) is the vertex. The parent function is \( y=x^2 \) (vertex \( (0,0) \), \( a = 1 \), opens up). The given graph in part (b) opens down, so \( a<0 \). Let's find two points. When \( x = 0 \), \( y = 3 \)? Wait, no, looking at the graph, when \( x = 0 \), the y - value is 3? Wait, the graph in part (b) has a vertex at \( (0,4) \)? Wait, maybe I made a mistake. Let's consider the reflection and vertical stretch. The parent function \( f(x)=x^2 \), if we reflect it over the x - axis, we get \( y=-x^2 \), which has vertex at \( (0,0) \) and opens down. But the given graph has a vertex at \( (0,4) \)? Wait, no, let's check the y - intercept. When \( x = 0 \), the graph in part (b) passes through \( (0,3) \)? Wait, maybe the correct approach is: The parent function \( y = x^2 \), the transformed function is \( y=-x^2+4 \)? Wait, let's test. If \( x = 0 \), \( y = 4 \), which matches the graph (the vertex is at \( (0,4) \)). So, first, reflection over the x - axis (so the coefficient of \( x^2 \) is negative) and then a vertical shift up by 4 units. The general form is \( y=-x^2 + 4 \). Let's verify with \( x = 1 \). For \( y=-x^2+4 \), when \( x = 1 \), \( y=-1 + 4=3 \). Looking at the graph, when \( x = 1 \), the y - value is 3 (approximately), which matches. So the equation is \( y=-x^2+4 \). Wait, but maybe the stretch? Wait, the parent function \( y=x^2 \) at \( x = 1 \) has \( y = 1 \). The reflected function \( y=-x^2 \) at \( x = 1 \) has \( y=-1 \). The given graph at \( x = 1 \) has \( y = 3 \), so \( y=-x^2+4 \) at \( x = 1 \) is \( y = 3 \), which is correct. So the equation is \( y=-x^2 + 4 \).

Answer:

(a) \( y=\boxed{x^2 - 3} \)
(b) \( y=\boxed{-x^2 + 4} \)