Given:
The table of values of a linear relationship.
x y
-1 -1
0 1
1 3
2 5
To find:
The equation for the given table of values.
Solution:
If a linear function passes through the two points, then the equation of the linear relationship is
![y-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)](https://img.qammunity.org/2022/formulas/mathematics/high-school/wkwv82bw6qlga765myohf3n6p3g9tbbqs4.png)
Consider any two point from the given table. Let the two points are (-1,-1) and (0,1). So, the equation of the linear relationship is
![y-(-1)=(1-(-1))/(0-(-1))(x-(-1))](https://img.qammunity.org/2022/formulas/mathematics/high-school/6l4dwmbfdc804ads5emj4bv5fna8melev2.png)
![y+1=(1+1)/(0+1)(x+1)](https://img.qammunity.org/2022/formulas/mathematics/high-school/2pq9z9umi5id9osqxadh3vilahwpzjk5e4.png)
![y+1=(2)/(1)(x+1)](https://img.qammunity.org/2022/formulas/mathematics/high-school/6g6cn6x5ygut4trew0dfjl3lyto9d7u7qk.png)
![y+1=2(x+1)](https://img.qammunity.org/2022/formulas/mathematics/high-school/ehmy453rkuivgbe7uoedk1gx02mrtswup0.png)
Using distributive property, we get
![y+1=2(x)+2(1)](https://img.qammunity.org/2022/formulas/mathematics/high-school/dxpn23ntrdyrmutjvetvlv6zwsaw6qdb9s.png)
![y+1=2x+2](https://img.qammunity.org/2022/formulas/mathematics/high-school/o1ko80u8vkby5tg42443nl5kyt1n66d115.png)
Subtracting 1 from both sides, we get
![y+1-1=2x+2-1](https://img.qammunity.org/2022/formulas/mathematics/high-school/uk07zhft09yafmj76f1teo6w276co1uon8.png)
![y=2x+1](https://img.qammunity.org/2022/formulas/mathematics/college/u7c6cirh4j6j52wal591t25fj5gwaq37tv.png)
Therefore, the required equation is
. Hence, the correct option is D.