$f(t) = 4t^2 - 16t + t$. Opening upward parabola. Find the number of integer values of $t$ for which $f(1) > 0$, $f(2) < 0$, $f(3) > 0$.
Step-by-Step Solution
Key Concept: For a quadratic function, use the given conditions at specific points to establish intervals on the parameter, then count integers in the intersection.
We have $f(t) = 4t^2 - 16t + t = 4t^2 - 15t$. From the graph and conditions: $f(1) > 0 \Rightarrow 4 - 16 + t > 0 \Rightarrow t > 12$. $f(2) < 0 \Rightarrow 16 - 32 + t < 0 \Rightarrow t < 16$. $f(3) > 0 \Rightarrow 36 - 48 + t > 0 \Rightarrow t > 12$. Therefore $12 < t < 16$, giving integer values $t \in \{13, 14, 15\}$.
Correct Answer: 3