Q. 9 CYU - Algebra 1

Question

The time in seconds that it takes an object to fall a distance d is given by the function $\mathit{t}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\sqrt{\mathbf{d}}$ (assuming zero air resistance). Graph the function , and state the domain and range.

Step-by-Step Solution

Answer

The domain of the given function is $\mathit{d}\mathbf{\in }\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$ and the range is $\mathit{t}\mathbf{\in }\mathbf{[}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{)}$.

Step 1. State the concept of parent graph.

Parent graph: The simplest form of the given function is called the parent function of that function and the graph of the parent function is called parent graph.

Step 2. State the concept of domain and range.

Domain: The set of all possible values for which given function defined is called domain.

Range: The set of all possible values of the given function is called range.

Step 3. Graph the function.

The given function is: $t=\frac{1}{4}\sqrt{d}$

In order to graph a function, find few co-ordinates by substituting values of ‘$d$’ and find finding the respective values of ‘$t$’. 

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}d=0,\\ t=\frac{1}{4}\sqrt{0}\\ =\frac{1}{4}\left(0\right)\\ =0\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}d=1,\\ t=\frac{1}{4}\sqrt{1}\\ =\frac{1}{4}\left(1\right)\\ =0.25\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}d=4,\\ t=\frac{1}{4}\sqrt{4}\\ =\frac{1}{4}\left(2\right)\\ =0.5\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}d=9,\\ t=\frac{1}{4}\sqrt{9}\\ =\frac{1}{4}\left(3\right)\\ =0.75\end{array}$

$\begin{array}{c}\text{For\hspace{0.17em}\hspace{0.17em}}d=16,\\ t=\frac{1}{4}\sqrt{16}\\ =\frac{1}{4}\left(4\right)\\ =1\end{array}$

Values of distance ‘$d$’	Values of   time ‘$t$’ in secs	$\left(d,t\right)$
0	0	(0,0)
1	0.25	(1,0.25)
4	0.5	(4,0.5)
9	0.75	(9,0.75)
16	1	(16,1)

Plot these co-ordinates on a coordinate plane, plotting the values of distance ‘$d$’ on $x$-axis and the values of time ‘$t$’ on $y$-axis and join those points to get the required graph.

Step 4. State the domain and range.

Since ‘$d$’ is inside the root, the values inside the root must be positive. 

Therefore, values of $d$ is all positive real numbers including zero. 

That is, $d\ge 0,\Rightarrow d\in [0,\infty )$. 

Therefore, domain: $[0,\infty )$

In $t=\frac{1}{4}\sqrt{d}$, the square root of $d$  is multiplied by $\frac{1}{4}$.

As square root is always positive, the least value of ‘$d$’  is zero. 

Therefore, find the least value of the time(t) by substituting in $t=\frac{1}{4}\sqrt{d}$.

$\begin{array}{c}t=\frac{1}{4}\left(\sqrt{0}\right)\\ =\frac{1}{4}\left(0\right)\\ =0\end{array}$ 

Since the coefficient of $\sqrt{d}$ is positive, time ‘$t$’ takes all the values in real numbers which are greater than or equal to zero.

Therefore, $t\ge 0,\Rightarrow t\in [0,\infty )$$t\ge 0,\Rightarrow t\in [0,\infty )$

Therefore, Range: $[0,\infty )$