We need to find a function of the form:
[tex]a\mleft|x-b\mright|+c[/tex]Such that it equals 6 when x=0 and 0 when x=5.
Substitute x=5 and assume that the expression is equal to 0:
[tex]a\mleft|5-b\mright|+c=0[/tex]Since the vertex of the function is at the point (5,0), we can assume that c=0.
Therefore:
[tex]a\mleft|5-b\mright|=0[/tex]Divide both sides by a:
[tex]|5-b|=0[/tex]Using that for any number k, |k| = 0 if and only if k=0, then:
[tex]5-b=0[/tex][tex]\text{Therefore, b=5.}[/tex]Next, substitute x=0 and assume that the expression is equal to 6 to find a.
[tex]a\mleft|0-5\mright|=6[/tex]Since |0-5|=5:
[tex]5a=6[/tex]Dividing both sides by 5, we get the value of a:
[tex]a=\frac{6}{5}[/tex]Substitute a=6/5, b=5 and c=0 in the original equation:
[tex]\frac{6}{5}|x-5|[/tex]You can check that this expression is equal to 6 when x=0 and it is equal to 0 when x=5.
To find the height at which the second studend should catch the ball, notice that the second student will be placed at x=5+4=9.
Substitute x=9 to find how high will the ball bounce for the second student to catch it:
[tex]\frac{6}{5}|9-5|=\frac{6}{5}|4|=\frac{6\cdot4}{5}=\frac{24}{5}=4.8\text{ ft}[/tex]To summarize:
[tex]\begin{gathered} \text{The function which represents this situation is:} \\ f(x)=\frac{6}{5}|x-5| \\ \\ \text{The ball bounces up to }4.8\text{ feet for the second student to catch it.} \end{gathered}[/tex]