Question

input output 3 6 5 7

Explanation:

Step1: Identify the graph type

The graph has a parabola (quadratic) and a line. We need to find the output for input \( x = 3 \) using the line or the parabola? Wait, the table has input 6 with output 5, which matches the parabola? Wait, no, let's check the line. The line passes through, let's find its equation. Let's take two points on the line. When \( x = 1 \), \( y = 0 \)? Wait, no, looking at the graph, the blue line (straight) passes through, maybe \( (1, 0) \) and \( (0, -5) \)? Wait, slope \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{0 - (-5)}{1 - 0}=5 \). So equation \( y = 5x - 5 \). Let's test \( x = 3 \): \( y = 5(3)-5 = 10 \)? Wait, no, maybe the parabola? Wait, the table has \( x = 6 \), output 5. Let's check the parabola: vertex at \( (5, 8) \)? Wait, no, the parabola has roots at \( x = 1 \) and \( x = 7 \)? Wait, no, the parabola crosses x-axis at \( x = 1 \) and \( x = 7 \)? Wait, no, looking at the graph, the parabola (purple) has vertex at \( (5, 8) \), and roots at \( x = 1 \) and \( x = 9 \)? Wait, maybe I'm confused. Wait, the input 6 has output 5. Let's check the line: if \( x = 6 \), \( y = 5(6)-5 = 25 \), no. Wait, the table says input 6, output 5. So maybe the parabola. Let's find the equation of the parabola. Vertex form: \( y = a(x - h)^2 + k \). Vertex at \( (5, 8) \), so \( y = a(x - 5)^2 + 8 \). It passes through \( (6, 5) \): \( 5 = a(6 - 5)^2 + 8 \Rightarrow 5 = a + 8 \Rightarrow a = -3 \). So equation \( y = -3(x - 5)^2 + 8 \). Now, for \( x = 3 \): \( y = -3(3 - 5)^2 + 8 = -3(4) + 8 = -12 + 8 = -4 \)? No, that doesn't match. Wait, maybe the line is the one we need. Wait, the blue line: when \( x = 1 \), \( y = 0 \); when \( x = 0 \), \( y = -5 \). So slope \( m = 5 \), equation \( y = 5x - 5 \). Now, \( x = 3 \): \( y = 5(3) - 5 = 10 \)? Wait, but the graph: when \( x = 3 \), the line is at \( y = 10 \)? Wait, maybe I made a mistake. Wait, the table has input 6, output 5. Let's check the line at \( x = 6 \): \( y = 5(6)-5 = 25 \), no. The parabola at \( x = 6 \): \( y = -3(1)^2 + 8 = 5 \), which matches the table. So the parabola is \( y = -3(x - 5)^2 + 8 \). Now, for \( x = 3 \): \( y = -3(3 - 5)^2 + 8 = -3(4) + 8 = -12 + 8 = -4 \)? No, that can't be. Wait, maybe the line is the function for the table? Wait, the table has input 3, output? Let's look at the graph again. The straight line (blue) passes through \( (1, 0) \) and \( (0, -5) \), so slope 5, equation \( y = 5x - 5 \). Let's check \( x = 3 \): \( y = 5*3 - 5 = 10 \). Wait, but the table has \( x = 6 \), output 5, which doesn't match the line. So maybe the table is for the parabola. Let's recheck: parabola equation \( y = -3(x - 5)^2 + 8 \). For \( x = 6 \), \( y = -3(1) + 8 = 5 \), correct. For \( x = 3 \), \( y = -3(9) + 8 = -27 + 8 = -19 \)? No, that's not right. Wait, maybe the vertex is at \( (5, 9) \). Then \( y = a(x - 5)^2 + 9 \). At \( x = 6 \), \( 5 = a(1)^2 + 9 \Rightarrow a = -4 \). Then \( x = 3 \): \( y = -4(4) + 9 = -16 + 9 = -7 \). No. Wait, maybe I'm looking at the wrong graph. Wait, the input is 3, let's find the y-value on the line (blue) at \( x = 3 \). From the graph, when \( x = 3 \), the blue line (straight) is at \( y = 10 \)? Wait, the grid: each square is 1 unit. So x=3, y=10? Wait, the table has input 6, output 5. Let's check the parabola at x=6: y=5, which is correct. So the line is another function. Wait, maybe the problem is to find the output for x=3 using the line. Let's confirm the line's equation. When x=1, y=0; x=0, y=-5. So slope is (0 - (-5))/(1 - 0)=5. Equation: y=5x -5. So for x=3, y=5*3 -5=10. So the output for input 3 is 10.

Step2: Calculate the output

Using the line's equation \( y = 5x - 5 \), substitute \( x = 3 \):
\( y = 5(3) - 5 = 15 - 5 = 10 \).

Answer:

10