Answer:
the zeros of the function are: 9 and 4
Step-by-step explanation:
Given the function:
[tex]f(t) = t^2-13t+36[/tex]
To find the zeros of the function:
Set f(t) = 0
then;
[tex]t^2-13t+36 = 0[/tex]
Split the middle term as: -9 and -4 we have;
[tex]t^2-9t-4t+36 = 0[/tex]
⇒[tex]t(t-9)-4(t-9) = 0[/tex]
Take (t-9) common we have;
[tex](t-9)(t-4) = 0[/tex]
By zero product property we have;
t-9 = 0 and t-4 = 0
⇒t = 9 and t = 4
Therefore, the zeros of the function are: 9 and 4
