Answer:
![\text{ y = -}(1)/(2)(t^2-7t+12)](https://img.qammunity.org/2023/formulas/mathematics/high-school/p0idx51cmh5hs6w7mbb5mi3ngv63tx8ncr.png)
Step-by-step explanation:
Here, we want to write the equation of the quadratic function
We have the general form as:
![\text{ y = ax}^2\text{ + bx + c}](https://img.qammunity.org/2023/formulas/mathematics/college/og4idotdnmlzy0prs4rad9v46t8dkv1ffy.png)
from the question, we have the horizontal intercepts at t= 3 and t =4
that means (t-3) is a factor and also (t-4) is another root of the equation
The product of these two is as follows:
![\text{ a(t-3)(t-4) = a(t}^2-7t\text{ + 12)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/cg9wrmbrduifnk7d0ufyto5tygk18bwkao.png)
Finally, we need to find the value of a
We can do this by making a substitution
We substitute 1 for t and -3 on the other side of the equation
Mathematically, we have this as:
![\begin{gathered} \text{ a(1-7+12) = -3} \\ 6a\text{ = -3} \\ a\text{ = }(-3)/(6)\text{ = -}(1)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/wu287wbk84vmddi1swhn26quwnhh0r7zwb.png)
Thus, we have the equation as:
![\text{ y = -}(1)/(2)(t^2-7t+12)](https://img.qammunity.org/2023/formulas/mathematics/high-school/p0idx51cmh5hs6w7mbb5mi3ngv63tx8ncr.png)