How do you graph #y = 2 sqrt (x) + 3# by plotting point?

1 Answer

First look at the basic shape of the graph, understand the modifiers, then find plot points that are more straightforward to solve for to get the graph below:


When being asked to graph something by plotting points, I find it is easier to do it if you have an idea as to what, in general, the shape of the graph is going to be (plotting a line is so much easier when you know it's a line you are plotting!)

So - what is the basic shape? While the 2 times and the plus 3 will modify it a bit, it's the #sqrtx# that is going to determine the shape. And what does that look like?

graph{sqrtx [-3.875, 16.125, -2.44, 7.56]}

The graph starts at the origin (remember that we can't graph the square root of negative numbers on this graph). It will increase in value into infinity (it never levels off) but does slow down how fast it rises significantly as x increases.

Now to the modifiers. The times 2 will make the y values twice as large for each x. And the plus 3 means the graph starts at #(0,3)# and has the same shape from there.

Lastly, let's get a couple of points plotted that will help. The key to finding plot points is finding the ones that are easy to solve and graph.

So for #y=2sqrtx+3#, there are a few graph points that suggest themselves. For instance:

#x=0# - I love using this one - the x term just disappears and you are left with something simple. In this case, #y=3#. So we have (0,3).

Since this is a square root, let's use perfect squares as plot points. We can use #x=1#, which gives #y=4#

#x=4#, which gives #y=7#

#x=9#, which gives #y=9#

(This is so much easier than trying to plot #x=2# and #y=2sqrt2+3#!)

That will give the final graph of:

graph{2sqrtx+3 [-2.17, 15, -1, 10]}