Question

If f(t) = 4t² − 3t + 6, find (i) f(0), (ii) f(4), (iii) f(−5).

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)f\left(t\right)=4t^2-3t+6 \\ \Rightarrow f\left(0\right)=\left(4\times 0^2-3\times 0+6\right) \end{gathered}$$
$$\begin{gathered} =\left(0-0+6\right) \\ =6 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)f\left(t\right)=4t^2-3t+6 \\ \Rightarrow f\left(4\right)=\left(4\times 4^2-3\times 4+6\right) \end{gathered}$$
$$\begin{gathered} =\left(64-12+6\right) \\ =58 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)f\left(t\right)=4t^2-3t+6 \\ \Rightarrow f\left(-5\right)=\left[4\times {\left(-5\right)}^2-3\times \left(-5\right)+6\right] \end{gathered}$$
$$\begin{gathered} =\left(100+15+6\right) \\ =121 \end{gathered}$$