Question

which table represents a quadratic function? \\(\

$$\begin{array}{|c|c|c|c|c|} \\hline x & -2 & -1 & 0 & 1 & 2 \\\\ \\hline f(x) & -9 & -4.5 & 0 & 4.5 & 9 \\\\ \\hline \\end{array}$$

\\) \\(\

$$\begin{array}{|c|c|c|c|c|} \\hline x & -2 & -1 & 0 & 1 & 2 \\\\ \\hline f(x) & 6 & 3 & 2 & 3 & 6 \\\\ \\hline \\end{array}$$

\\) \\(\

$$\begin{array}{|c|c|c|c|c|} \\hline x & -2 & -1 & 0 & 1 & 2 \\\\ \\hline f(x) & -5 & -3 & -1 & 1 & 3 \\\\ \\hline \\end{array}$$

\\) \\(\

$$\begin{array}{|c|c|c|c|c|} \\hline x & -2 & -1 & 0 & 1 & 2 \\\\ \\hline f(x) & 0.25 & 0.5 & 1 & \\text{(incomplete)} & \\text{(incomplete)} \\\\ \\hline \\end{array}$$

\\)

Explanation:

Step1: Recall quadratic function test

A quadratic function has constant second differences of $f(x)$ when $x$-values are equally spaced.

Step2: Test Table 1 (top-left)

$x$: $-2, -1, 0$; $f(x)$: $0.25, 0.5, 1$
First differences: $0.5-0.25=0.25$, $1-0.5=0.5$
Second difference: $0.5-0.25=0.25$ (only 1 second diff, not enough, and first diffs grow exponentially, so it's exponential)

Step3: Test Table 2 (top-right)

$x$: $-2, -1, 0, 1, 2$; $f(x)$: $-9, -4.5, 0, 4.5, 9$
First differences: $-4.5-(-9)=4.5$, $0-(-4.5)=4.5$, $4.5-0=4.5$, $9-4.5=4.5$
Second differences: $4.5-4.5=0$ (constant zero, so it's linear)

Step4: Test Table 3 (middle-right)

$x$: $-2, -1, 0, 1, 2$; $f(x)$: $6, 3, 2, 3, 6$
First differences: $3-6=-3$, $2-3=-1$, $3-2=1$, $6-3=3$
Second differences: $-1-(-3)=2$, $1-(-1)=2$, $3-1=2$ (constant second difference of 2)

Step5: Test Table 4 (bottom-right)

$x$: $-2, -1, 0, 1, 2$; $f(x)$: $-5, -3, -1, 1, 3$
First differences: $-3-(-5)=2$, $-1-(-3)=2$, $1-(-1)=2$, $3-1=2$
Second differences: $2-2=0$ (constant zero, so it's linear)

Answer:

The table with $x$ values $-2, -1, 0, 1, 2$ and corresponding $f(x)$ values $6, 3, 2, 3, 6$ (middle-right table) represents a quadratic function.