Answer
a) $3500$ and $6875$; the savings deposited in 1995 and 2010.
b) 2025
c) $\$225/year$
Work Step by Step
a) Plugging in $t=0,15$ into the given function, we get:
$D(0)=3500+15(0)^2=3500$
$D(15)=3500+15(15)^2=6875$
These values represent the savings deposited in 1995 and 2010.
b) To find the year in which Ella deposits $\$17,000$, we plug in $\$17,000$ for $D$ and solve for $t$:
$3500+15t^2=17000\\13500=15t^2\\900=t^2\\t=\pm30$
But $t$ must be positive, so we choose $t=+30$, which represents the year $2025$.
c) The rate of change is:
$\frac{f(y)-f(x)}{y-x}=\frac{6875-3500}{15-0}=\$225/year$
This value represents the average increase in deposit per.
