Question:
If $f(t)=4 t^{2}-3 t+6$, find
(i) $f(0)$,
(ii) $f(4)$,
(iii) $f(-5)$.
Solution:
(i) $f(t)=4 t^{2}-3 t+6$
$\Rightarrow f(0)=\left(4 \times 0^{2}-3 \times 0+6\right)$
$=(0-0+6)$
$=6$
(ii) $f(t)=4 t^{2}-3 t+6$
$\Rightarrow f(4)=\left(4 \times 4^{2}-3 \times 4+6\right)$
$=(64-12+6)$
$=58$
(iii) $f(t)=4 t^{2}-3 t+6$
$\Rightarrow f(-5)=\left[4 \times(-5)^{2}-3 \times(-5)+6\right]$
$=(100+15+6)$
$=121$
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