### 4.1 Messing with the Basic Graph

Lesson materials located below the video overview.

Consider the squaring function $$y=x^{2}$$ which takes an input $$x$$ and squares it to produce the output $$x^{2}$$. For example:

If $$x=1$$, then $$y=1^{2}=1$$.

If $$x=5$$, then $$y=5^{2}=25$$.

If $$x=-5$$, then $$y=(-5)^{2}=25$$.

If $$x=20$$, then $$y=20^{2}=400$$.

If $$x=0$$, then $$y=0^{2}=0$$.

And so on.

If we draw a table and plot points, we see that the graph of $$y=x^{2}$$ is an upward facing U-shaped curve.

The important point:  THE FUNCTION $$y=x^{2}$$ GIVES A SYMMETRICAL U-SHAPED GRAPH.

CURIOUS QUESTION: Is “U-shape” an accurate term? The sides of the letter U look vertical. Do the sides of $$y=x^{2}$$ ever become perfectly vertical? If we accurately plot the graph of $$y=x^{2}$$, to what extent is it actually U like?

[Pushing this idea … Suppose we rotate the graph of about the origin $$0.03$$ degrees counter-clockwise. Does the rotated graph cross the $$y$$-axis above the origin? What do you think?]

Here is a challenge question for students, to be asked right now, without anything but the graph of under our belts!

I am six feet tall and am standing at the position $$x=4$$ feet on the horizontal axis. Is it possible to write down a formula for a function whose graph is the same U-shaped curve as for $$y=x^{2}$$ but positioned to balance on my head?

Perhaps try this before reading on. Play with some possible formulae. Plug in some sample points and table values. Test whether your ideas offer any hints as to a solution to this challenge.

The COMPANION GUIDE to this QUADRATICS course contains an essay on how I actually conduct this exercise in my classes.

### CHANGE 1: A NEW ZERO FOR THE $$x$$-VALUES

Let’s now be systematic with our changes to the graph.

Comment: It is much better to discover all these changes for yourself by battling with the previous exercise of balancing a U-shape curve on my head! This is the trouble with textbook writing: one has to give all the answers away!

Here is the graph of $$y=x^{2}$$ again:

It has a table of values:

One thing to notice is that this graph has a “dip” at $$x=0$$.

Now consider the function $$y=(x-3)^{2}$$.

Notice that when we put $$x=3$$ into this formula we obtain the output $$0^{2}$$. That is, the number $$3$$ is “behaving” just like $$x=0$$ was for the original function.

In $$y=(x-3)^{2}$$ we have that $$3$$ is the “new zero” for the $$x$$-values.

So whatever the original function was doing at $$x=0$$, it is now doing it at $$x=3$$.

The original function $$y=x^2$$ dips at $$x=0$$,  so the graph of the function $$y=(x-3)^{2}$$ dips at $$x=3$$.

The entire graph has been shifted horizontally – and one can check this by drawing a table of values if you like:

EXAMPLE 56: Sketch a graph of  $$y=(x+3)^{2}$$.

Answer: What value of $$x$$ behaves like zero for $$y=(x+3)^{2}$$? Answer: $$x=-3$$ does!

So $$y=(x+3)^{2}$$ looks like $$y=x^{2}$$ but with $$x=-3$$ the new zero.

Recall that solutions to all practice problems appear in the COMPANION GUIDE to this QUADRATICS course.

PRACTICE 57:

a) Sketch a graph of $$y=(x-4)^{2}$$.

b) Sketch a graph of $$y=(x+\frac{1}{2})^{2}$$.

### CHANGE 2: SHIFTING UPWARDS

Here’s the function $$y=x^{2}$$ again:

How would the graph of $$y=x^{2}+3$$ appear?

Notice that this new function is adds three units to each output:

This has the effect of raising the entire graphs three units in the vertical direction:

The graph of the function $$y=x^{2}-5$$ would be the same graph shifted downwards $$5$$ units, and the graph of $$y=x^{2}+\sqrt{3}$$  would the graph shifted upwards $$\sqrt{3}$$ units.

PRACTICE 58:

a) Sketch $$y=x^{2}-5$$

b) Sketch $$y=x^{2}+\frac{1}{2}$$.

### CHANGE 3: BOTH TOGETHER

Here’s the graph of $$y=x^{2}$$:

and here is the graph of $$y=(x-2)^{2}$$.  Here “$$2$$” is acting as the new zero for the $$x$$-values, so the dip that was occurring at zero is now occurring at $$2$$:

Now consider $$y=(x-2)^{2}+3$$ . This is the previous graph shifted upwards three units:

Question: Here we went from $$y=x^{2}$$ to $$y=(x-2)^{2}$$ to $$y=(x-2)^{2}+3$$ drawing the three graphs above along the way. Is it possible to think, instead, of the sequence “from $$y=x^{2}$$ to $$y=x^{2}+3$$ to $$y=(x-2)^{2}+3$$”?

Comment:  Some people find it easiest just to follow the standard “order of operations” one learns in algebra: Deal with the effect of the innermost parentheses first, and then work out from there. They would thus “construct” $$y=(x-2)^{2}+3$$ via the chain “$$y=x^{2}$$ to $$y=(x-2)^{2}$$ to $$y=(x-2)^{2}+3$$.”

PRACTICE 59:

a) Sketch $$y=(x-4)^{2}-1$$.

b) Sketch $$y=(x+4)^{2}+2$$.

c) Sketch $$y=(x+1)^{2}-2$$.

PRACTICE 60: I am six feet tall and am standing at the position $$x=4$$ feet on the horizontal axis. Write down the formula of a U-shaped graph that sits balanced on my head.

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