Question

Find the domain and sketch the graph of the functions\({\bf{H}}\left( {\bf{t}} \right){\bf{ = }}\frac{{{\bf{4 - }}{{\bf{t}}^{\bf{2}}}}}{{{\bf{2 - t}}}}\).

Short answer

The domain of the function\({\bf{H}}\left( {\bf{t}} \right){\bf{ = }}\frac{{{\bf{4 - }}{{\bf{t}}^{\bf{2}}}}}{{{\bf{2 - t}}}}\)is\(\left( { - \infty ,2} \right) \cup \left( {2,\infty } \right)\)and the graph of the function is given in figure (1).

Step-by-step solution

Determine the domain of the function

The given function\(H\left( t \right) = \frac{{4 - {t^2}}}{{2 - t}}\)holds true for all the real numbers except for the numbers for which the denominator becomes zero. The denominator of the function becomes zero for numbers,

\(\begin{aligned}{c}2 - t = 0\\t = 2\end{aligned}\)

Therefore, the function is defined for\(t \in \left( { - \infty ,2} \right) \cup \left( {2,\infty } \right)\).

Thus, the domain of the function \(H\left( t \right) = \frac{{4 - {t^2}}}{{2 - t}}\)is\(\left( { - \infty ,2} \right) \cup \left( {2,\infty } \right)\).

Sketch the graph of the function

Now take\(t\)on the\(x - \)axis, and\(H\left( t \right)\)on the\(y - \)axis.

Simplify the function.

\(\begin{aligned}{c}H\left( t \right) &= \frac{{4 - {t^2}}}{{2 - t}}\\H\left( t \right) &= \frac{{\left( {2 + t} \right)\left( {2 - t} \right)}}{{2 - t}}\\H\left( t \right) &= 2 + t\end{aligned}\)

For\(t = 0\)

\(\begin{aligned}{l}H\left( t \right) &= 2 + 0\\H\left( t \right) &= 2\end{aligned}\)

For \(H\left( t \right) = 0\)

\(\begin{aligned}{c}0 = 2 + t\\t = - 2\end{aligned}\)

Therefore, the graph passes through (-2, 0) and (0, 2).

Draw the graph of the function using the above two points.

Figure (1)

Therefore, the graph of the function is given in figure (1).