$$a(t)=Zt^2+Bt+1$$
If \$100 at $t=0$ grows to \$152 at $t=4$ and \$200 at $t=0$ grows to \$240 at $t=2$, what are $Z$ and $B$? Please show work. Also, what would \$1600 invested at $t=6$ grow to at $t=8$?
$$a(t)=Zt^2+Bt+1$$
If \$100 at $t=0$ grows to \$152 at $t=4$ and \$200 at $t=0$ grows to \$240 at $t=2$, what are $Z$ and $B$? Please show work. Also, what would \$1600 invested at $t=6$ grow to at $t=8$?
On
Part A
$\bullet$ consider the amount function $a(t)=\dfrac{A(t)}{A(0)}$,
$\bullet$ use the information given as follows to solve for $Z$ and $B$
$A(0)=100,A(4)=152$
and
$A(0)=200,A(2)=240$
Part B
for an amount invested at $s>0$ the value of the investment at $t>s$ is
$A(t)=A(s)\dfrac{a(t)}{a(s)}$
with $s=6,t=8$
$A(8)=1600\dfrac{a(8)}{a(6)}$
Since $\$ 100$ grows to $\$152$ from time $t=0$ to $t=4$, we know that $a(4) = 1.52$. Likewise, we know from the information about the $\$ 200$ investment that $a(2) = 1.2$. This is simply using the definition of the accumulation function.
Now, we have two equations and two unknowns:
$a(2) = 1.2 = 2^2 Z + 2B + 1$
$a(4) = 1.52 = 4^2 Z+4B+1$
Solve these equations using elimination or substitution.
Let me know if you need more help.