Question

museum visitors
which equation best models the graph?
a. $n = -2x^2 + 20x + 58$
b. $n = -2x^2 + 25x + 78$
c. $n = 2x^2 + 20x + 58$
d. $n = 2x^2 + 20x$

Explanation:

Step1: Analyze the parabola's direction

The graph of museum visitors first increases and then decreases, so the parabola opens downward. For a quadratic equation \(N = ax^2+bx + c\), if \(a<0\), the parabola opens downward. So we can eliminate options C (\(a = 2>0\)) and D (\(a = 2>0\)) since their leading coefficients are positive (parabola opens upward).

Step2: Test a point in the remaining options

Let's take \(x = 0\) (Day 0). From the graph, when \(x = 0\), \(N\approx80\).
- For option A: \(N=-2(0)^2 + 20(0)+58=58\). But \(58 eq80\), so A is incorrect.
- For option B: \(N=-2(0)^2+25(0)+78 = 78\). Close to 80 (considering graph approximation). Let's test another point, say \(x = 1\). From the graph, \(x = 1\) (Day 1) has \(N\approx76\).
For option B: \(N=-2(1)^2+25(1)+78=-2 + 25+78=101\)? Wait, no, wait, maybe I misread the graph. Wait, when \(x = 0\), the point is around 80. Wait, maybe \(x = 0\) is \(N = 80\) (the first point). Wait, let's recalculate option B at \(x = 0\): \(N=-2(0)^2+25(0)+78 = 78\), which is close to 80. Let's check \(x = 2\). From the graph, \(x = 2\) (Day 2) has \(N = 90\). For option B: \(N=-2(2)^2+25(2)+78=-8 + 50+78=120\)? No, that's not right. Wait, maybe I made a mistake. Wait, the equation is \(N=-2x^2+25x + 78\)? Wait, no, maybe the graph's \(x\) is day, let's check \(x = 1\). The point at \(x = 1\) (Day 1) is around 76. Let's check option B: \(x = 1\), \(N=-2(1)^2+25(1)+78=-2 + 25+78 = 101\)? That's not matching. Wait, maybe I messed up the first step. Wait, the parabola opens downward, so \(a<0\), so A and B have \(a=-2\). Wait, let's check option A at \(x = 2\). Option A: \(N=-2(2)^2+20(2)+58=-8 + 40+58=90\). From the graph, at \(x = 2\) (Day 2), \(N = 90\) (the point is at 90). Yes! So for \(x = 2\), option A gives \(N = 90\), which matches the graph. Let's check \(x = 0\) again. Option A: \(N = 58\), but the graph at \(x = 0\) is 80. Wait, maybe the \(x = 0\) point is not exactly 80, maybe the graph's \(x = 0\) is the first point, but maybe the equation is better fitted. Wait, let's check \(x = 3\). For option A: \(N=-2(3)^2+20(3)+58=-18 + 60+58=100\). From the graph, at \(x = 3\) (Day 3), the point is around 100 - 105? Wait, the point at \(x = 3\) is 104? Wait, no, the graph: Day 0: ~80, Day 1: ~76, Day 2: 90, Day 3: 104? Wait, no, the red dots: first dot (x=0) is ~80, x=1: ~76, x=2: 90, x=3: 104? Wait, maybe my initial analysis of \(x = 0\) was wrong. Let's use \(x = 2\). Option A: \(x = 2\), \(N=-2(4)+40 + 58=-8 + 40+58=90\), which matches the graph (Day 2 has 90 visitors). Option B: \(x = 2\), \(N=-2(4)+50 + 78=-8 + 50+78=120\), which is way off. So option A is better. Wait, earlier when \(x = 0\), option A gives 58, but maybe the graph's \(x = 0\) is a bit different. Alternatively, maybe the first point is \(x = 0\), \(N = 80\), but the equation is an approximation. Since options C and D are eliminated (open upward), and between A and B, when \(x = 2\), option A gives \(N = 90\) (matches the graph), while option B does not. So option A is correct? Wait, but when \(x = 0\), option A gives 58, but maybe the graph's \(x = 0\) is not the first point. Wait, the graph has points at \(x = 0\) (N=80), \(x = 1\) (N=76), \(x = 2\) (N=90), \(x = 3\) (N=104), etc. Wait, let's check \(x = 3\) for option A: \(N=-2(9)+60 + 58=-18 + 60+58=100\), which is close to 104. For option B: \(x = 3\), \(N=-2(9)+75 + 78=-18 + 75+78=135\), which is too high. So option A is better. Wait, maybe my initial \(x = 0\) calculation was wrong. The equation \(N=-2x^2+20x + 58\) at \(x = 0\) is 58, but the graph's \(x = 0\) is 80. Wait, maybe the \(x\) - axis is day starting from 0, but the first point is \(x = 0\), \(N = 80\). But maybe the model is an approximation. Since the parabola opens downward (so \(a<0\), eliminating C and D), and option A fits better at \(x = 2\) (N=90) and \(x = 3\) (N≈100), while option B gives too high values. So the correct answer is A? Wait, no, wait, when \(x = 0\), option A is 58, but the graph is 80. Maybe I made a mistake in the point. Wait, maybe the first point is \(x = 0\), \(N = 58\)? No, the graph shows \(x = 0\) as around 80. Wait, maybe the equation is \(N=-2x^2+20x + 58\). Let's check \(x = 5\). From the graph, \(x = 5\) (Day 5) has \(N\approx110\). For option A: \(N=-2(25)+100 + 58=-50 + 100+58=108\), close to 110. For option B: \(x = 5\), \(N=-2(25)+125 + 78=-50 + 125+78=153\), which is way off. So option A is better.

Answer:

A. \(N = -2x^2 + 20x + 58\)