Question

which graph represents the function $f(x) = -(x - 3)^2 + 1$?
options: a) graph with vertex at (-3,1), opening downward, crossing x-axis at -4 and -2
b) graph with vertex at (3,1), opening upward

Explanation:

Step1: Identify Vertex Form

The function \( f(x) = -(x - 3)^2 + 1 \) is in vertex form \( f(x) = a(x - h)^2 + k \), where \((h, k)\) is the vertex. Here, \( h = 3 \), \( k = 1 \), so the vertex is \((3, 1)\). Also, \( a = -1 \), which means the parabola opens downward (since \( a < 0 \)).

Step2: Analyze Graphs

- Graph a: Vertex at \((-3, 1)\), opens downward.
- Graph b: Vertex at \((3, 1)\), opens upward (since it's a U - shape). Wait, no—wait, graph b: let's check the direction. Wait, the coefficient \( a=-1 \), so it should open downward. Wait, no, graph b: the parabola in b is opening upward (since it's a minimum at \((3,1)\)), but our function has \( a = -1 \), so it should open downward. Wait, no, maybe I misread. Wait, the function is \( -(x - 3)^2+1 \), so \( a=-1 \), so it's a downward - opening parabola with vertex at \((3,1)\). Wait, looking at the graphs:

Wait, graph a: vertex at \((-3,1)\), opens downward. Graph b: vertex at \((3,1)\), but opens upward? Wait, no, maybe the options are mis - presented? Wait, no, the user's graph: let's re - check. Wait, the function is \( f(x)=-(x - 3)^2 + 1 \). So vertex at \((3,1)\), opens downward. Wait, but in the given options, maybe there's a mistake? Wait, no, maybe I made a mistake. Wait, the original problem: the two graphs. Let's check the vertex coordinates.

Graph a: x - coordinate of vertex is - 3, y - coordinate is 1. Graph b: x - coordinate of vertex is 3, y - coordinate is 1. Now, the coefficient \( a=-1 \), so the parabola opens downward. So the graph with vertex at \((3,1)\) and opening downward. But in the given options, maybe the correct graph is not shown? Wait, no, maybe I misread the function. Wait, the function is \( -(x - 3)^2+1 \), so vertex at \((3,1)\), opens downward. So which graph has vertex at \((3,1)\) and opens downward? Wait, in the given options, graph b: let's check the direction. Wait, graph b: the parabola in b is a U - shape (opens upward), but our function should open downward. Wait, maybe there's a typo, but according to the vertex form, the vertex is \((3,1)\) and it opens downward. Wait, maybe the correct graph is not among a and b? No, the user provided two graphs. Wait, maybe I made a mistake in the direction. Wait, \( a=-1 \), so it's a downward - opening parabola. So the graph with vertex at \((3,1)\) and opening downward. But in the given options, graph b has vertex at \((3,1)\) but opens upward, graph a has vertex at \((-3,1)\) and opens downward. Wait, this is a problem. Wait, no, maybe the function is \( -(x + 3)^2+1 \)? No, the function is \( -(x - 3)^2+1 \). Wait, maybe the user's graph is mis - labeled. But according to the vertex form, the vertex is \((3,1)\) and it opens downward. So if we assume that there's a graph with vertex at \((3,1)\) and opening downward, but in the given options, maybe the intended correct graph is the one with vertex at \((3,1)\) and opening downward. But in the given options, graph b has vertex at \((3,1)\) but opens upward, graph a has vertex at \((-3,1)\) and opens downward. This is a contradiction. Wait, maybe the original function was \( -(x + 3)^2+1 \), but it's written as \( -(x - 3)^2+1 \). If it's \( -(x + 3)^2+1 \), then vertex at \((-3,1)\) and opens downward, which is graph a. But the function is \( -(x - 3)^2+1 \). Wait, maybe there's a mistake in the problem. But according to the standard vertex form, the vertex is \((h,k)\) where \( h = 3 \), \( k = 1 \), and opens downward. So the correct graph should have vertex at \((3,1)\) and open downward. But in the given options, if we have to choose between a and b, and assuming that maybe there's a typo in the function (maybe \( -(x + 3)^2+1 \)), then graph a is correct. But according to the given function, the vertex is \((3,1)\). Wait, maybe the user's graph for b is actually opening downward? Let's re - look at the graph for b: the graph in b has a minimum at \((3,1)\), so it opens upward. The graph in a has a maximum at \((-3,1)\), so it opens downward. So according to the function \( f(x)=-(x - 3)^2+1 \), the vertex is \((3,1)\) and it should open downward. But there's no such graph. Wait, maybe the function is \( (x - 3)^2+1 \) (without the negative sign), then it would open upward and graph b would be correct. But the function has a negative sign. This is confusing. But maybe the intended answer is that the graph with vertex at \((3,1)\) and opening downward, but since graph b has vertex at \((3,1)\) (even though it's opening upward, maybe a mistake), or graph a has vertex at \((-3,1)\) (if the function was \( -(x + 3)^2+1 \)). Given that the function is \( -(x - 3)^2+1 \), the vertex is \((3,1)\) and it opens downward. So if we assume that there's a graph with vertex at \((3,1)\) and opening downward, but in the given options, maybe the correct one is the one with vertex at \((3,1)\), so graph b (even though it's opening upward, maybe a mistake in the graph's direction). Or maybe I made a mistake. Wait, let's calculate some points. For \( f(x)=-(x - 3)^2+1 \):

When \( x = 3 \), \( f(3)=1 \).

When \( x = 2 \), \( f(2)=-(2 - 3)^2+1=-1 + 1 = 0 \).

When \( x = 4 \), \( f(4)=-(4 - 3)^2+1=-1 + 1 = 0 \).

So the parabola passes through \((2,0)\), \((3,1)\), \((4,0)\) and opens downward. Now, looking at graph b: does it pass through \((2,0)\), \((3,1)\), \((4,0)\)? Graph b: at \( x = 3 \), \( y = 1 \), at \( x = 2 \), \( y \) is 2? No, wait, graph b's grid: let's see, the x - axis is from - 5 to 5, y - axis from - 4 to 5. At \( x = 3 \), \( y = 1 \). At \( x = 2 \), \( y = 2 \)? No, that doesn't match. At \( x = 2 \), \( f(2)=0 \), so the graph should pass through \((2,0)\) and \((4,0)\). Graph a: at \( x=-4 \), \( y = 0 \); \( x=-2 \), \( y = 0 \); vertex at \((-3,1)\). So if the function was \( -(x + 3)^2+1 \), then \( f(-4)=-(-4 + 3)^2+1=-1 + 1 = 0 \), \( f(-2)=-(-2 + 3)^2+1=-1 + 1 = 0 \), which matches graph a. So maybe there's a typo in the function, and it's \( -(x + 3)^2+1 \) instead of \( -(x - 3)^2+1 \). In that case, graph a is correct. Given that, we'll proceed with the assumption that maybe there's a typo and the function is \( -(x + 3)^2+1 \), so the graph with vertex at \((-3,1)\) and opening downward is graph a.

Answer:

a) The graph with vertex at \((-3, 1)\) and opening downward (assuming a typo in the function's horizontal shift, or a mis - drawn graph for the correct vertex position).