## Quadratics

### 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.

**YOUR TURN …**

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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