Quadratics

4.4 Common Sense Graphing – The Power of Symmetry

Lesson materials located below the video overview.

In the first part of this graphing section we saw that it is easy to graph quadratics of the type:

\(y=4(x-7)^{2}+9\).

This example is a fairly steep upward-facing U graph with \(x=7\) as the \(x\)-value of the vertex and whole graph is shifted upwards \(9\)  units. (So \(y=9\) is the \(y\)-coordinate of the vertex.)

 

In the second part of this section we proved:

The graph of \(y=ax^{2}+bx+c\) is a symmetrical U-shaped graph just like \(y=x^{2}\) . The steepness is \(a\).

So, for example, \(y=-2x^{2}-672x+109\) is a downward facing U graph.

 

But it is still hard to determine where the vertex of this graph lies, and so it is still hard to actually graph the curve.

 

However …

If we can find two interesting and symmetrical \(x\)-values, then common sense will point the way!

 

Let me explain what I mean by this.

 

EXAMPLE 76: Sketch \(y=2(x-3)(x-9)+7\).

 

Answer: The first thing to note is that if we were to expand \(2(x-3)(x-9)+7\) we would obtain an expression of the form \(ax^{2}+bx+c\) . (Actually, check this: Show that\(2(x-3)(x-9)+7\) expanded equals \(2x^{2}-24x+34\).)

 

Without doing all the expanding, however, we can see that the positive \(2\) in front will help give the term \(2x^{2}\) in the expansion. We have a positive steepness.

So we know \(y=2(x-3)(x-9)+7\) is going to give a symmetrical upward-facing U-shaped graph.

 

In \(y=2(x-3)(x-9)+7\) two obvious and interesting \(x\)-values are staring us in the face:

Put \(x=3\) and we get \(y=0+7=7\).

Put \(x=9\) and we get \(y=0+7=7\).

So we now have an upward-facing U that goes through two symmetrical points:

Q4_4_pic1

Since the graph is symmetrical, common sense says that the graph must have vertex at \(x=6\), the point halfway between \(x=3\) and \(x=9\).

The trouble is we don’t know how high the U is at this point: Does the U sit above the -axis? Dip below it? Just touch it?

Q4_4_pic2

 Well … let’s just plug \(x=6\) into the formula to find out!

When \(x=6\) we have \(y=2(3)(-3)+7=-11\).

We must have the following picture:

Q4_4_pic3

DONE!  □

NOTICE: I am personally not at all fussed about scale in the picture nor extra-fussy details about the graph. As long as the information given on the diagram is clear and correct and gives all the key features of interest, then all is good. (But watch out: Some curriculum writers are much more fussy about these details than me.)

 

 

EXAMPLE 77: Sketch \(y=-(x+2)(x-10)+1\).  Where is its vertex? What is its -intercept? What is the graph’s largest value?

 

Answer: It will be a symmetrical U-shaped curve, downward facing (the steepness is \(-1\)).

Also \(x=-2\) and \(x=10\) are interesting. They both give \(y=-0+1=1\).

Q4_4_pic5

The vertex lies at \(x=4\) and there \(y=-(6)(-6)+1=37\).

Q4_4_pic6

 

The vertex has coordinates \( \left( {4},{37} \right) \).  The largest value of the graph is \(y=37\).  And the \(y\)-intercept occurs when \(x=0\) which gives \(y=-(2)(-10)+1=21\). □

 

In these two examples interesting \(x\)-values were laid bare.

HOW DOES ONE FIND INTERESTING x-VALUES IF THEY ARE NOT OBVIOUS?

Let’s address this with an example:

 

Consider \(y=x^{2}+4x+5\).  This is going to be an upward-facing U-shaped graph.

 

To find interesting \(x\)-values, focus on the “\(x\) part” of the formula, in this case, just:  \(x^{2}+4x\).

 

The only thing I can think to do with is to rewrite it as \(x(x+4)\) and so:

\(y=x(x+4)+5\).

 

Now it is clear that \(x=0\) and \(x=-4\) are interesting. They both give \(y=5\).

 

By symmetry, we know the vertex lies midway between these interesting \(x\)-values, at \(x=-2\), and here \(y=(-2)(2)+5=1\).

Here then is its graph.

Q4_4_pic7

 

EXAMPLE 78: Make a sketch of \(y=-3x^{2}+6x+7\). Rewrite this equation in “vertex form” (whatever that means!)

 

Answer: This is a downward facing U.

Focusing on the \(x\) part, from \(-3x^{2}+6x\) we see we can factor \(3x\). We have:

\(y=3x(-x+2)+7\).

It is now clear that \(x=0\) and \(x=2\) are interesting. They each give \(y=7\).

The vertex must occur halfway between these two values, at \(x=1\). Here, \(y=-3+6+7=10\).

This is enough to give a pretty good sketch!

Q4_4_pic8

 

Now we have to make an intelligent guess as to what “vertex form” means.

The vertex of this U is \(\left({1},{10}\right)\) so we know we could also write this curve as:

\(y=a(x-1)^{2}+10\)

for some steepness  \(a\). This is probably what “vertex form” means.

To find \(a\) let’s plug in a point, say, \(x=0\), \(y=7\):

\(7=a(-1)^{2}+10\)

\(a=-3\)

So we have \(y=-3(x-1)^{2}+10\).   □

 

EXAMPLE 79: Make a sketch of \(y=-2x^{2}+3x+7\). What are its \(x\)-and \(y\)-intercepts?

 

Answer: This is a downward facing U-shaped curve. We have:

\(y=-2x^{2}+3x+7=x(-2x+3)+7\)

and this U has the value \(7\) at both \(x=0\) and \(x=\frac{3}{2}\). Because the graph is symmetrical, the vertex must be halfway between these values, at \(x=\frac{3}{4}\). At this value, \(y=\frac{3}{4}\left( -2 \cdot \frac{3}{4} + 3\right) + 7=8 \frac{1}{8}\).

The graph appears:

Q4_4_pic9

The \(y\)-intercept is clearly \(y=7\).

The \(x\)-intercepts occur where \(y=0\). We need to solve:

\(-2x^{2}+3x+7=0\)

The box method gives (CHECK THIS!):

\(x=\frac{3 \pm \sqrt{65}}{4}\).    □

 

EXAMPLE 80: Find a formula for the following U-shaped curve:

Q4_4_pic10

 

Answer: The picture shows us that \(x=17\) and \(x=27\) give symmetrical outputs. This suggests the formula:

\(y=a(x-17)(x-27)+13\).

We inserted a value \(a\) as we don’t yet know the steepness.

However, we are also told that when \(x=30\) we have \(y=60\). So we need:

\(60=a(13)(3)+13\)

\(47=39a\)

\(a=\frac{47}{39}\)

So \(y=\frac{47}{39} (x-17)(x-27) +13\) does the trick!  □

 

YOUR TURN …

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

 

PRACTICE 81: Make reasonable sketches of the following quadratics:

a) \(y-x^{2}-2x+3\)

b) \(y=4x^{2}+8x+5\)

c) \(y=-5x^{2}-20x+10\)

d) \(y=-2(x+30)(x-50)\)

 

PRACTICE 82: Make reasonable sketches of the following quadratics:

a) \(y=3(x-1)^{2}+5\)

b) \(y=(x-2)(x+3)+8\)

c) \(y=-5x^{2}-20x+10\)

d) \(y=7(x-2)(x+3)\)

 

PRACTICE 83: A quadratic crosses the \(x\)-axis at \(x=2\) and at \(x=8\),  and it crosses the \(y\)-axis at \(y=7\). Find a formula for the quadratic.

HINT: Make a sketch of the information you have so far and use your common sense to see what the quadratic must be doing.

 

PRACTICE 84: Find a formula for the following quadratic:

Q4_4_pic11

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