Question 1120055
A person makes annual payments of $ 1000 into an ordinary annuity.
 At the end of 5 years, the amount in the annuity is $ 5703.88.
 What annual nominal compounding rate has this annuity earned?
:
We can use the annuity formula
P({{{((1+i)^t-1)/i}}}) = Fv; where
P =payment
i = interest rate
t = periods
Fv = Final value
:
1000({{{((1+i)^5-1)/i}}}) = 5703.88
Put in the form that we can graph; i = x
1000({{{((1+x)^5-1)/x}}}) = 5703.88
divide both sides by 1000
{{{((1+x)^5-1)/x}}} = {{{5703.88/1000}}}
Approx solution; x = .066 or 6.6% interest
{{{((1+x)^5-1)/x}}} = 5.704
multiply both sides by x
(1+x)^5 - 1 = 5.704x
A quadratic equation we can graph
(1+x)^5 - 5.704x - 1 = 0
:
{{{ graph( 300, 200, -.1, .1, -.05, .05, (1+x)^5-5.704x-1) }}}