Question 1203719
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Assuming the four types of equations are in fact linear, quadratic, exponential, and square root....<br>
(1) exponential and square root functions are monotonically increasing where they are defined. Since the behavior must be y goes to infinity as x goes to negative infinity, the leftmost part of the graph must be either linear (with a negative slope) or quadratic.
(2) a linear equation is probably the easiest to fit with the other types of functions, so let's not start with a linear function.  So we will make the leftmost part of the graph quadratic.  {{{y=x^2}}} is a simple one to work with.
(3) (1,1) is a solution to both {{{y=x^2}}} and {{{y=sqrt(x)}}}, so it looks as if making the second function equal to {{{y=sqrt(x)}}} will be convenient. We can stop using that function at a convenient point, say (4,2).
(4) Making the third function linear will make it easy to make the "junction" between it and the fourth (exponential) function continuous.  So choose a simple linear function containing (4,2): {{{y=x-2}}}.  We can end it again at a convenient point, say (5,3).
(5) Then the fourth function is exponential containing (5,3).  Although it doesn't look "nice", the function {{{y=e^x+(3-e^5)}}} will do the job.<br>
One of an infinite number of possibilities....<br>
{{{y=x^2}}} for x less than 1 [red in the graph below]
{{{y=sqrt(x)}}} for x greater than or equal to 1 and less than 4 [green]
{{{y=x-2}}} for x greater than or equal to 4 and less than 5 [blue]
{{{y=e^x+(3-e^5)}}} for x greater than or equal to 5 [purple]<br>
Here is a graph with the four functions overlapping, and showing the piecewise graph continuous at (1,1), (4,2), and (5,3):<br>
{{{graph(400,400,-6,10,-6,10,x^2,sqrt(x),x-2,e^x+3-e^5)}}}<br>