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.

Q4_1_pic1

 

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?

Q4_1_pic2

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:

Q4_1_pic3

It has a table of values:

Q4_1_pic4

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 \(x=3\).

Q4_1_pic5

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

Q4_1_pic6

 

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.

Q4_1_pic7

 

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:

Q4_1_pic8

Q4_1_pic9

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

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

Q4_1_pic10

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

Q4_1_pic11

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 \(y=x^{2}\):

Q4_1_pic12

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 \(2\):

Q4_1_pic13

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

Q4_1_pic14

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.

Q4_1_pic15

 

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