Question

complex numbers
complex numbers review
0 0 5 10 x
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the parabola passes through coordinates (2, 8), (3, 5), (4, 4), (5, 5), and (6, 8).
which function shares no real solutions with the quadratic function in the graph?
(1 point)
○ $y = \frac{1}{2}x - 1$
○ $y = -\frac{1}{2}x + 6$
○ $y = x^2$
○ $y = x + 1$

Explanation:

Step1: Analyze the parabola's range

The parabola passes through points with \( y \)-values: \( 8, 5, 4, 5, 8 \). So the minimum \( y \)-value of the parabola is \( 4 \) (at \( x = 4 \)).

Step2: Analyze each linear function's range relative to the parabola

- For \( y=\frac{1}{2}x - 1 \): As \( x \) varies, \( y \) can be less than \( 4 \) (e.g., \( x = 0 \), \( y=-1 \)) and intersect the parabola? Wait, no—wait, the parabola's minimum \( y \) is \( 4 \). Let's check the linear functions' \( y \)-values at the parabola's \( x \)-range ( \( x \) from \( 2 \) to \( 6 \) ).
- For \( y = x^2 \): The parabola \( y = x^2 \) has a minimum \( y = 0 \) (at \( x = 0 \)), but our given parabola has minimum \( y = 4 \). Let's see if \( y = x^2 \) and the given parabola intersect. The given parabola at \( x = 2 \), \( y = 8 \); \( x^2 = 4 \) (not 8). At \( x = 3 \), \( y = 5 \); \( x^2 = 9 \) (not 5). At \( x = 4 \), \( y = 4 \); \( x^2 = 16 \) (not 4). At \( x = 5 \), \( y = 5 \); \( x^2 = 25 \) (not 5). At \( x = 6 \), \( y = 8 \); \( x^2 = 36 \) (not 8). Also, the given parabola's \( y \) is between \( 4 \) and \( 8 \) for \( x \) in \( [2,6] \), while \( y = x^2 \) at \( x \) in \( [2,6] \) has \( y \) from \( 4 \) to \( 36 \), but the given parabola's \( y \) is at most \( 8 \), so \( y = x^2 \) at \( x \) in \( [2,6] \) is \( \geq 4 \), but at \( x = 2 \), \( x^2 = 4 \), given parabola at \( x = 2 \) is \( 8 \); \( x = 3 \), \( x^2 = 9 \), given is \( 5 \); so \( y = x^2 \) is always above the given parabola? Wait, no—wait the given parabola's equation: let's find its equation. The parabola is symmetric about \( x = 4 \) (since \( x = 2 \) and \( x = 6 \) have \( y = 8 \); \( x = 3 \) and \( x = 5 \) have \( y = 5 \); vertex at \( (4,4) \). So equation is \( y = a(x - 4)^2 + 4 \). Plug in \( (2,8) \): \( 8 = a(2 - 4)^2 + 4 \Rightarrow 8 = 4a + 4 \Rightarrow 4a = 4 \Rightarrow a = 1 \). So equation is \( y = (x - 4)^2 + 4 = x^2 - 8x + 20 \).

Now, check intersections with each option:

1. \( y=\frac{1}{2}x - 1 \): Set \( x^2 - 8x + 20=\frac{1}{2}x - 1 \Rightarrow x^2 - \frac{17}{2}x + 21 = 0 \). Discriminant: \( (\frac{17}{2})^2 - 4 \times 1 \times 21 = \frac{289}{4} - 84 = \frac{289 - 336}{4}=-\frac{47}{4}<0 \)? Wait no, wait \( \frac{289}{4}=72.25 \), \( 84 = 84 \), so \( 72.25 - 84 = -11.75 \)? Wait, no, I miscalculated. Wait \( (17/2)^2 = 289/4 = 72.25 \), \( 4*1*21 = 84 \), so discriminant is \( 72.25 - 84 = -11.75 < 0 \)? But that can't be. Wait, maybe my parabola equation is wrong. Wait, the parabola passes through \( (2,8) \), \( (3,5) \), \( (4,4) \), \( (5,5) \), \( (6,8) \). So symmetric about \( x = 4 \), vertex at \( (4,4) \), so \( y = a(x - 4)^2 + 4 \). Plug \( (2,8) \): \( 8 = a( - 2)^2 + 4 \Rightarrow 8 = 4a + 4 \Rightarrow a = 1 \). So equation is \( y = (x - 4)^2 + 4 = x^2 - 8x + 20 \). Now, check \( y = \frac{1}{2}x - 1 \): \( x^2 - 8x + 20 = 0.5x - 1 \Rightarrow x^2 - 8.5x + 21 = 0 \). Discriminant: \( (8.5)^2 - 4*1*21 = 72.25 - 84 = -11.75 < 0 \)? No, wait 8.5 squared is 72.25, 4*21 is 84, so 72.25 - 84 = -11.75 < 0. But wait, the linear function \( y = \frac{1}{2}x - 1 \) at \( x = 10 \), \( y = 4 \), but our parabola at \( x = 10 \) isn't given, but our parabola is between \( x = 2 \) and \( x = 6 \). Wait, maybe I made a mistake. Let's check the other options.

2. \( y = -\frac{1}{2}x + 6 \): Set \( x^2 - 8x + 20 = -0.5x + 6 \Rightarrow x^2 - 7.5x + 14 = 0 \). Discriminant: \( (7.5)^2 - 4*1*14 = 56.25 - 56 = 0.25 > 0 \), so real solutions.

3. \( y = x^2 \): Set \( x^2 - 8x + 20 = x^2 \Rightarrow -8x + 20 = 0 \Rightarrow x = 20/8 = 2.5 \). Then \( y = (2.5)^2 = 6.25 \). Check if \( (2.5, 6.25) \) is on the given parabola: \( y = (2.5 - 4)^2 + 4 = (-1.5)^2 + 4 = 2.25 + 4 = 6.25 \). So they intersect at \( (2.5, 6.25) \).

4. \( y = x + 1 \): Set \( x^2 - 8x + 20 = x + 1 \Rightarrow x^2 - 9x + 19 = 0 \). Discriminant: \( 81 - 76 = 5 > 0 \), real solutions.

Wait, but earlier analysis was wrong. Wait the correct approach: the given parabola has vertex at \( (4,4) \), opening upwards (since \( a = 1 > 0 \)). The linear function \( y = \frac{1}{2}x - 1 \): let's find its \( y \)-value at \( x = 4 \) (vertex of parabola): \( y = \frac{1}{2}(4) - 1 = 2 - 1 = 1 \), which is below the parabola's \( y = 4 \) at \( x = 4 \). Now, check if the line and parabola intersect. The parabola's equation is \( y = (x - 4)^2 + 4 \), the line is \( y = 0.5x - 1 \). Subtract: \( (x - 4)^2 + 4 - (0.5x - 1) = (x^2 - 8x + 16) + 5 - 0.5x = x^2 - 8.5x + 21 \). Discriminant: \( (8.5)^2 - 4*1*21 = 72.25 - 84 = -11.75 < 0 \), so no real solutions. Wait, but the other options:

- \( y = -\frac{1}{2}x + 6 \): at \( x = 4 \), \( y = -\frac{1}{2}(4) + 6 = -2 + 6 = 4 \), which is equal to the parabola's \( y \) at \( x = 4 \). So they intersect at \( (4,4) \)? Wait no, the parabola at \( x = 4 \) is \( y = 4 \), and the line at \( x = 4 \) is \( y = 4 \), so that's a point of intersection. Wait, my earlier discriminant for \( y = -\frac{1}{2}x + 6 \) was wrong. Let's recalculate: \( x^2 - 8x + 20 = -0.5x + 6 \Rightarrow x^2 - 7.5x + 14 = 0 \). Discriminant: \( (7.5)^2 = 56.25 \), \( 4*1*14 = 56 \), so \( 56.25 - 56 = 0.25 \), so roots are \( x = \frac{7.5 \pm 0.5}{2} \), so \( x = 4 \) or \( x = 3.5 \). At \( x = 4 \), \( y = 4 \), which is the vertex of the given parabola. So they intersect.

- \( y = x + 1 \): at \( x = 3 \), \( y = 4 \); the given parabola at \( x = 3 \) is \( y = 5 \), so no. Wait, no, solving \( x^2 - 8x + 20 = x + 1 \Rightarrow x^2 - 9x + 19 = 0 \), roots \( x = \frac{9 \pm \sqrt{5}}{2} \approx \frac{9 \pm 2.236}{2} \), so \( x \approx 5.618 \) or \( x \approx 3.382 \), which are in \( [2,6] \), so they intersect.

- \( y = x^2 \): we found intersection at \( x = 2.5 \), which is in \( [2,6] \), so they intersect.

- \( y = \frac{1}{2}x - 1 \): discriminant negative, so no real solutions. Wait, but at \( x = 10 \), the line \( y = \frac{1}{2}(10) - 1 = 4 \), and the given parabola at \( x = 10 \) isn't given, but the given parabola is only defined (from the points) between \( x = 2 \) and \( x = 6 \). Wait, but the problem says "the quadratic function in the graph"—the graph's parabola is between \( x = 2 \) to \( x = 6 \), but the linear functions are defined for all \( x \). However, the key is which function has no real solutions (i.e., no intersection points) with the quadratic.

Wait, maybe I messed up the parabola's equation. Let's rederive it. The parabola is symmetric about \( x = 4 \) (since \( x = 2 \) and \( x = 6 \) have \( y = 8 \); \( x = 3 \) and \( x = 5 \) have \( y = 5 \); vertex at \( (4,4) \). So equation is \( y = a(x - 4)^2 + 4 \). Plug in \( (3,5) \): \( 5 = a(3 - 4)^2 + 4 \Rightarrow 5 = a + 4 \Rightarrow a = 1 \). So equation is \( y = (x - 4)^2 + 4 = x^2 - 8x + 20 \), correct.

Now, check \( y = \frac{1}{2}x - 1 \):

Set \( x^2 - 8x + 20 = \frac{1}{2}x - 1 \)

\( x^2 - \frac{17}{2}x + 21 = 0 \)

Multiply by 2: \( 2x^2 - 17x + 42 = 0 \)

Discriminant: \( (-17)^2 - 4*2*42 = 289 - 336 = -47 < 0 \)

Ah! I forgot to multiply by 2 earlier. So discriminant is negative, meaning no real solutions.

For \( y = -\frac{1}{2}x + 6 \):

\( x^2 - 8x + 20 = -\frac{1}{2}x + 6 \)

\( x^2 - \frac{15}{2}x + 14 = 0 \)

Multiply by 2: \( 2x^2 - 15x + 28 = 0 \)

Discriminant: \( (-15)^2 - 4*2*28 = 225 - 224 = 1 > 0 \), so real solutions ( \( x = \frac{15 \pm 1}{4} = 4 \) or \( 3.5 \) ), which we saw earlier.

For \( y = x^2 \):

\( x^2 - 8x + 20 = x^2 \Rightarrow -8x + 20 = 0 \Rightarrow x = 2.5 \), real solution.

For \( y = x + 1 \):

\( x^2 - 8x + 20 = x + 1 \Rightarrow x^2 - 9x + 19 = 0 \)

Discriminant: \( 81 - 76 = 5 > 0 \), real solutions.

So the function with no real solutions is \( y = \frac{1}{2}x - 1 \).

Step3: Confirm the answer

After calculating discriminants for each function's intersection with the quadratic, only \( y = \frac{1}{2}x - 1 \) has a negative discriminant, meaning no real solutions.

Answer:

\( y = \frac{1}{2}x - 1 \) (the first option: \( \boldsymbol{y = \frac{1}{2}x - 1} \))