Answer: The correct option is (D)
[tex]y=\begin{array}{cc}\{ &\begin{array}{cc}x^2-4 & x\leq 1\\x^2+3 & x>1\end{array}\end{array}[/tex]
Step-by-step explanation: We are given to select the correct function that is graphed in the figure.
We have,
[tex]\textup{If }y=x^2-4,\textup{ then }y(1)=1^2-4=1-4=-3,\\\\\textup{If }y=x^2+3,\textup{ then }y(1)=1^2+3=1+3=4.[/tex]
In the graph, there is closed hole at y = - 3, so y = - 3 (x = 1) is included.
Also, there is an open hole at y = 4, so y = 4 (x = 1) is excluded.
So, If y = x² - 4, then x will be greater than or equal to 1.
If y = x² + 3, then x will be less than 1.
Also,
[tex]\textup{for }x\leq 1,~~y\leq x^2-4,\\\\\textup{for }x> 1,~~y> x^2+3.[/tex]
Therefore, the graphed function will be
[tex]y=\begin{array}{cc}\{ &\begin{array}{cc}x^2-4 & x\leq 1\\x^2+3 & x>1\end{array}\end{array}[/tex]
Option (D) is the correct function.
