Question

this table shows a relationship between x and y.

x	y
2	10
3	24

write an equation to represent the relationship.

Explanation:

Step1: Check linear relationship

Calculate the slope between (1,8) and (2,10): $m_1=\frac{10 - 8}{2 - 1}=2$. Between (2,10) and (3,24): $m_2=\frac{24 - 10}{3 - 2}=14$. Not linear.

Step2: Check non - linear (product - based)

For $x = 1,y = 8=1\times8$; $x = 2,y = 10$ (not $2\times8$). Wait, another approach: Let's see if $y$ is related to factorial or other. Wait, $1\times8 = 8$, $2\times5=10$? No. Wait, $8 = 1\times8$, $10=2\times5$, $24 = 3\times8$? No. Wait, maybe $y=8x + 2(x - 1)(x - 2)$? Wait, no. Wait, let's check the differences. First differences: 10 - 8 = 2, 24 - 10 = 14. Second differences: 14 - 2 = 12. Not quadratic. Wait, maybe a piece - wise? No, the problem probably expects a pattern. Wait, when $x = 1$, $y = 8$; $x = 2$, $y = 10$ (8+2); $x = 3$, $y = 24$ (10 + 14). No. Wait, maybe $y=8x+2(x - 1)$? For $x = 1$: $8\times1+2\times0 = 8$. $x = 2$: $8\times2+2\times1=16 + 2=18
eq10$. No. Wait, another way: Let's see the product of $x$ and something. Wait, $x = 1$, $y = 8$; $x = 2$, $y = 10$ (8 + 2); $x = 3$, $y = 24$ (10+14). Wait, maybe the pattern is $y = 8x+2(x - 1)(x - 2)$? No. Wait, maybe I made a mistake. Wait, let's list the values:

$x = 1,y = 8$

$x = 2,y = 10$

$x = 3,y = 24$

Wait, another approach: Let's assume the equation is $y=ax^{2}+bx + c$.

For $x = 1$: $a + b + c=8$

For $x = 2$: $4a+2b + c = 10$

For $x = 3$: $9a+3b + c=24$

Subtract first from second: $3a + b=2$ (Equation 1)

Subtract second from third: $5a + b = 14$ (Equation 2)

Subtract Equation 1 from Equation 2: $2a=12\Rightarrow a = 6$

Then from Equation 1: $3\times6 + b=2\Rightarrow b=2 - 18=-16$

From first equation: $6-16 + c=8\Rightarrow c=18$

So the quadratic equation is $y = 6x^{2}-16x + 18$

Let's check:

For $x = 1$: $6(1)^{2}-16(1)+18=6 - 16 + 18 = 8$. Correct.

For $x = 2$: $6(4)-16(2)+18=24 - 32 + 18 = 10$. Correct.

For $x = 3$: $6(9)-16(3)+18=54 - 48 + 18 = 24$. Correct.

But maybe the problem has a simpler pattern. Wait, maybe I misread the table. Wait, if the third $y$ value was 24, and the first two are 8 and 10. Wait, another way: Let's see $y=8x + 2(x - 1)$? No. Wait, the quadratic equation we found works.

Answer:

$y = 6x^{2}-16x + 18$

Wait, but maybe there is a simpler pattern. Wait, let's check again. Wait, when $x = 1$, $y = 8$; $x = 2$, $y = 10$ (8 + 2); $x = 3$, $y = 24$ (10+14). The differences between $y$ values: 2,14. The difference of differences: 12. So the quadratic formula is $y=ax^{2}+bx + c$. We solved it and got $y = 6x^{2}-16x + 18$. Alternatively, maybe the problem has a typo, but based on the given values, the quadratic equation is correct.