Question

this table represents a quadratic function.

x	y
1	10.5
2	8
3	6.5
4	6
5	6.5

what is the value of a in the functions equation?

a. 2
b. \\(\frac{1}{2}\\)
c. 1
d. \\(-\frac{1}{2}\\)

Explanation:

Step1: Identify vertex form

A quadratic function in vertex form is \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. From the table, the vertex is at \( x = 4 \) (since \( y \) is symmetric around \( x = 4 \), as \( y(3)=6.5 \), \( y(5)=6.5 \); \( y(2)=8 \), \( y(6) \) would be 8, etc.), so \( h = 4 \), \( k = 6 \). So the equation is \( y = a(x - 4)^2 + 6 \).

Step2: Substitute a point

Use the point \( (0, 14) \) (when \( x = 0 \), \( y = 14 \)) into the equation:
\( 14 = a(0 - 4)^2 + 6 \)
Simplify: \( 14 = 16a + 6 \)

Step3: Solve for \( a \)

Subtract 6 from both sides: \( 8 = 16a \)
Divide by 16: \( a = \frac{8}{16} = \frac{1}{2} \)? Wait, no—wait, check symmetry again. Wait, the vertex is at \( x = 4 \), but let's check another point. Wait, when \( x = 2 \), \( y = 8 \). Let's use \( (2, 8) \):
\( 8 = a(2 - 4)^2 + 6 \)
\( 8 = 4a + 6 \)
Subtract 6: \( 2 = 4a \)
\( a = \frac{2}{4} = \frac{1}{2} \)? No, wait, maybe I made a mistake. Wait, the table: at \( x = 0 \), \( y = 14 \); \( x = 1 \), \( 10.5 \); \( x = 2 \), 8; \( x = 3 \), 6.5; \( x = 4 \), 6; \( x = 5 \), 6.5. Wait, the parabola opens upwards? But when \( x \) moves from 0 to 4, \( y \) decreases, then from 4 to 5, \( y \) increases. So vertex at \( (4, 6) \), so it opens upwards? But let's check the second differences. For a quadratic, the second differences are constant. Let's compute first differences (Δy):
From \( x=0 \) to \( x=1 \): \( 10.5 - 14 = -3.5 \)
\( x=1 \) to \( x=2 \): \( 8 - 10.5 = -2.5 \)
\( x=2 \) to \( x=3 \): \( 6.5 - 8 = -1.5 \)
\( x=3 \) to \( x=4 \): \( 6 - 6.5 = -0.5 \)
\( x=4 \) to \( x=5 \): \( 6.5 - 6 = 0.5 \)

Now second differences (Δ(Δy)):
\( -2.5 - (-3.5) = 1 \)
\( -1.5 - (-2.5) = 1 \)
\( -0.5 - (-1.5) = 1 \)
\( 0.5 - (-0.5) = 1 \)

For a quadratic function, the second difference is \( 2a \) (since the general form is \( y = ax^2 + bx + c \), and the second difference is \( 2a \times \Delta x^2 \); here \( \Delta x = 1 \), so second difference is \( 2a \)). Wait, the second difference is 1, so \( 2a = 1 \)? No, wait, the second difference for \( ax^2 \) with \( \Delta x = 1 \) is \( 2a \). Wait, but in our case, the second difference is positive 1, so \( 2a = 1 \) → \( a = \frac{1}{2} \)? But wait, the answer options include \( \frac{1}{2} \) (option B) and \( -\frac{1}{2} \) (option D). Wait, maybe I messed up the direction. Wait, let's re-express the vertex form. Wait, the vertex is at \( (4, 6) \), so \( y = a(x - 4)^2 + 6 \). Let's use \( x = 0 \), \( y = 14 \):
\( 14 = a(16) + 6 \) → \( 16a = 8 \) → \( a = 0.5 \), which is \( \frac{1}{2} \). But wait, the first differences: from \( x=0 \) to \( x=1 \), Δy = -3.5; \( x=1 \) to \( x=2 \), Δy = -2.5; so the differences are increasing (becoming less negative), which means the parabola opens upwards, so \( a \) is positive. So \( a = \frac{1}{2} \), which is option B. Wait, but let me check again with \( x = 1 \):
\( y = \frac{1}{2}(1 - 4)^2 + 6 = \frac{1}{2}(9) + 6 = 4.5 + 6 = 10.5 \), which matches. \( x = 2 \): \( \frac{1}{2}(4) + 6 = 2 + 6 = 8 \), matches. \( x = 3 \): \( \frac{1}{2}(1) + 6 = 0.5 + 6 = 6.5 \), matches. \( x = 5 \): \( \frac{1}{2}(1) + 6 = 6.5 \), matches. So \( a = \frac{1}{2} \), which is option B.

Answer:

B. \( \frac{1}{2} \)