For the following situations write a formula that gives the value of the investment, C, in terms of the number of years since the investment was made, x.The investment started with $700 and increased by $36 each year. The investment started with $1100 and increases by 1.46% each year. The investment started with $900 and decreases by 1.17% each year. The investment started with $1200 and decreases by $62 each year.

To find: The formula that gives the value of the investment, C in each situation. 1) Given: The investment started with $700 and increased by $36 each year. The investment increased by the same amount each year so it is a linear function. The slope is 36 and the y-intercept is 700. The investment after x years is given by, C(x)=36x+700. 2) Given: The investment started with $1100 and increases by 1.46% each year. After 1 year the investment becomes, C(1)=1100+1100×1.46100=1100(1+0.0146)=1100(1.0146) After 2 years the investment becomes, C(2)=1100(1.0146)+1100(1.0146)×1.46100=1100(1.0146)(1+0.0146)=1100(1.0146)(1.0146)=1100(1.0146)2 Therefore, The investment after x years is given by C(x)=1100(1.0146)x.3) Given: The investment started with $900 and decreases by 1.17% each year. After 1 year the investment becomes, C(1)=900-900×1.17100=900(1-0.0117)=900(0.9883) After 2 years the investment becomes, C(2)=900(0.9883)-900(0.9883)×1.17100=900(0.9883)(1-0.0117)=900(0.9883)(0.9883)=900(0.9883)2 Therefore, the investment after x years is given by C(x)=900(0.9883)x.4) Given: The investment started with $1200 and decreases by $62 each year. The investment decreased by the same amount each year so it is a linear function. The slope is -62 (negative because its decreasing) and the y-intercept is 1200. The investment after x years is given by, C(x)=-62x+1200.