### 4.3 A Necessary Piece of Theory

Lesson materials located below the video overview.

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

Here is the graph of the line $$y=x$$:

What do we obtain if we “add” these two formulas together and graph $$y=x^{2}+x$$?

Here are the two graphs on the same set of axes.

To the right, it is clear that “ $$x^{2}$$ plus $$x$$” adds together two positive numbers: The graph of $$y=x^{2}+x$$ is clearly getting larger and larger as we go over to the right.

Matters are less clear to the far left: The $$x^{2}$$ values are large and positive but the $$x$$ values are large and negative. Does the positive or the negative win? Will the graph of $$y=x^{2}+x$$ be positive to the left? Will it be negative to the left? Will it oscillate between the positives and negatives? What does it do?

The answer comes from going back to basics and just plotting points. Maybe try this on your own first before reading on.

Here is a table of values:

To be clear of what is happening between $$x=-1$$ and $$x=0$$, let’s try a few more values:

At $$x=-\frac{1}{2}$$ we have $$y=\left(-\frac{1}{2}\right) ^{2}+\left(-\frac{1}{2}\right) =-\frac{1}{4}=-0/25$$.

At $$x=-\frac{3}{4}$$ we have $$y=\frac{9}{16}-\frac{3}{4}=-\frac{3}{16}=-0.1875$$.

At $$x=-\frac{1}{4}$$ we have $$y=\frac{1}{16} -\frac{1}{4} = -\frac{3}{16} = -0/1875$$

The graph seems to dip down to lowest value $$-\frac{1}{4}$$ at $$x=-\frac{1}{2}$$.

Is this a perfect U-shaped graph, with $$x=-\frac{1}{2}$$ behaving as zero for the $$x$$-values and shift $$-\frac{1}{4}$$ units? That is, could we have $$y=a\left(x+\frac{1}{2}\right)^{2}-\frac{1}{4}$$ for some steepness $$a$$?

ACTUALLY … If  $$y=a\left(x+\frac{1}{2}\right)^{2}-\frac{1}{4}$$   goes through the point $$\left( {0},{0}\right)$$ then we need $$0=\frac{1}{4}a-\frac{1}{4}$$ giving $$a=1$$.

So is $$y=x^{2}+x$$ really the U-shaped graph $$y=a\left(x+\frac{1}{2}\right)^{2}-\frac{1}{4}$$ in disguise?

To answer that question let’s expand $$\left(x+\frac{1}{2}\right)^{2}$$:

$$\left( x+\frac{1}{2}\right)^{2}=x^{2}+\frac{1}{2}x + \frac{1}{2}x +\frac{1}{4}=x^{2}+x+\frac{1}{4}$$

and so

$$\left( x+\frac{1}{2}\right)^{2}-\frac{1}{4}=x^{2}+x+\frac{1}{4}=\frac{1}{4}=x^{2}+x$$.

YES!  $$y=x^{2}+x$$ is a perfect U-shaped quadratic graph with vertex at $$\left( {-\frac{1}{2}},{-\frac{1}{4}} \right)$$.

ANOTHER EXAMPLE:

Consider the graph of $$y=x^{2}+4x-3$$. Is this too a $$y=x^{2}$$ graph messed around with?

Let’s apply the box method to this formula.

We have an $$x^{2}$$ piece that comes from $$x \times x$$ and we have $$4x$$ which splits into two equal parts without fractions:

The box says we need the number $$4$$, not $$3$$, as part of the equation. Let’s add $$7$$ to each side:

$$y+7=x^{2}+4x+4$$

$$y+7=(x+2)^{2}$$

Now we see that

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

and so $$y=x^{2}+4x-3$$ is indeed a U-shaped graph with $$x=-2$$ the new zero for the $$x$$-values with the whole graph shifted downwards $$7$$ places.

A THIRD EXAMPLE:

Let’s make this a complicated one!

Is $$y=-3x^{2}-x+1$$ a $$y=x^{2}$$ formula messed around with?

Let’s apply the box method. Multiply through by $$-3$$ and then through by $$4$$:

$$-3y=9x^{2}+3x-3$$

$$-12y=36x^{2}+12x-12$$

Now work with the box:

The box shows we need to add $$13$$ to each side of the equation:

$$-12y+13=36x^{2}+12x+1$$

$$-12y+13=(6x+1)^{2}$$

It is not going to be pretty, but let’s pull out a factor of $$6$$ from the right hand term to make the variable $$x$$ stand proudly alone:

$$(6x+1)=6\left( x+\frac{1}{6}\right)$$

so

$$(6x+1)^{2}=36\left( x+\frac{1}{6}\right)^{2}$$

So right now we have:

$$-12y+13=36\left(x+\frac{1}{6}\right)^{2}$$.

Let’s make this an equation about $$y$$:

$$-12y=36\left( x+\frac{1}{6}\right)^{2}-13$$

so

$$y=-3\left( x+\frac{1}{6}\right)^{2} + \frac{13}{12}$$.

So now it is clear that even $$y=-3x^{2}-x+1$$ is the equation $$y=x^{2}$$ messed around with!

Its graph is an upside-down U-shaped curve with $$\left( -{\frac{1}{6}},{\frac{13}{12}}\right)$$ as the location of its vertex.

COMMENT: Notice that the original equation $$y=-3x^{2}-x+1$$ has the number $$-3$$ in front of the squared term, and also $$y=-3\left( x+\frac{1}{6}\right)^{2} + \frac{13}{12}$$ has the number $$-3$$ in front of its squared term. And this makes sense: If we were to expand $$y=-3\left( x+\frac{1}{6}\right)^{2} + \frac{13}{12}$$ only one $$x^{2}$$ term would appear and the number in front of it must be $$-3$$ in order to match the equation $$y=-3x^{2}-x+1$$.

In either formulation:

$$y=-3x^{2}-x+1$$ or $$y=-3\left( x+\frac{1}{6}\right)^{2} + \frac{13}{12}$$ the number in front of the squared term is the “steepness” of the U-curve.

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

PRACTICE 74: Repeat the above work to show that $$y=2x^{2} +8x+5$$ is a U-shaped curve. What is its steepness? Where is its vertex?

OPTIONAL PRACTICE 75: THIS IS FOR THE BRAVE

Show that any equation of the form $$y=ax^{2}+bx+c$$ is a version of $$Y=x^{2}$$ “messed around with.” What is the steepness of the U? Where is the vertex of the U?

SOME HINTS FOR THIS OPTIONAL PRACTICE QUESTION:

As before, use the box method. Start by multiplying through by $$4a$$:

$$4ay=4a^{2}x^{2}+4abx+4ac$$

Draw the box:

and use the box to get $$4ay-4ac+b^{2}=(2ax+b)^{2}$$.

Rewrite $$(2ax+b)$$ as $$2a\left( a+\frac{b}{2a}\right)$$ and see if you can then show:

$$y=a\left(x+\frac{b}{2a}\right)^{2} +c -\frac{b^{2}}{4a}$$.

This is U-shaped graph with steepness $$a$$. One can write down the coordinates of the vertex (but it is ugly).

The optional practice exercise proves the following result – the only thing we need from this section:

Every quadratic $$y=ax^{2}+bx+c$$ is a version of $$y=x^{2}$$ “messed around with”.

It equals $$a\left( x+ something\right)^{2} + something$$ with steepness $$a$$.

As such, the graph of $$y=ax^{2}+bx+c$$ is also a symmetrical U-shaped curve (with steepness $$a$$).

This is the key observation that is going to make graphing all quadratics ridiculously easy!

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