### 4.5 Some Traditional Textbook Questions

Lesson materials located below the video overview.

Let’s launch right one!

EXAMPLE 85: Here is a traditional algebra II question:

Consider $$y=2x^{2}-8x+6$$.

a) Find its vertex

b) Find its “line of symmetry” (whatever that means!)

c) Rewrite the quadratic in “vertex form.”

d) Sketch a graph of the quadratic

e) Find where the quadratic crosses the -axis.

f) Find where the quadratic crosses the -axis.

NO ONE SAYS YOU NEED ANSWER QUESTIONS IN THE ORDER GIVEN TO YOU!

And my second piece is:

ALWAYS DRAW A PICTURE FIRST – and then all the rest will be clear.

So let’s answer part d) first.

Interesting $$x$$-values:

$$y=2x(x-4)+6$$.

We see that $$x=0$$ and $$x=4$$ both give $$y=6$$.

The vertex must be at $$x=2$$, and here $$y=2(2)(-2)+6=-2$$.

We have:

Here are the easy parts to answer:

a) The vertex is at $$\left( {2},{-2}\right)$$.

b) The “line of symmetry” must mean the vertical line of symmetry at position $$2$$ on the $$x$$-axis. This vertical line has equation: $$x=2$$.

f) The $$y$$-intercept is $$y=6$$.

To answer c) we need to make use of the vertex to rewrite the equation.

The equation is:

$$y=a(x-2)^{2}-2$$

for some steepness $$a$$. (Actually… Since the formula is $$y=2x^{2}-8x+6$$ we know the steepness is $$a=2$$, but to check …) Let’s put in $$x=0$$, $$y=6$$ to find $$a$$:

$$6=a(-2)^{2}-2$$

$$8=4a$$

$$a=2$$

So the answer to part c) is:

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

To answer e) we need to put $$y=0$$ and solve:

$$0=2(x-2)^{2}-2$$

COMMENT: We could work with the original equation instead and solve $$0=2x^{2}-8x+6$$, but I suspect what I have chosen to do will be swifter. No big deal – all correct paths lead to the same correct results!

So

$$2(x-2)^{2}=2$$

$$(x-2)^{2}=1$$

$$x-2=1$$  or  $$x-2=-1$$

$$x=3$$ or $$x=1$$.

The $$x$$-intercepts are $$x=1$$ and $$x=3$$.  □

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

PRACTICE 86: Consider $$y=5x^{2}-10x-15$$.

a) Find its vertex

b) Find its “line of symmetry.”

c) Rewrite the quadratic in “vertex form.”

d) Sketch a graph of the quadratic

e) Find where the quadratic crosses the -axis.

f) Find where the quadratic crosses the -axis.

g) What is the smallest value of the graph?

### SUMMARISING THOUGHTS

We have seen that every quadratic equation $$y=ax^{2}+bx+c$$ is a symmetric U-shaped graph, namely the shape of $$y=x^{2}$$, messed around with.

There are two good ways to SEE the graph of a quadratic:

I: Is it obviously $$y=x^{2}$$ messed around with directly?

For example, $$y=-3(x+10)^{2}+17$$ is an upside-down U with $$x=-10$$ behaving like zero for the $$x$$-values and the graph is shifted upwards $$17$$ units. (The vertex of the graph is $$\left({-10},{17}\right)$$.)

II: Are there any interesting $$x$$-values?

For example, $$y=2(x+5)(x-17)+3$$ has two obvious symmetrical points: Both $$x=-5$$ and $$x=17$$ give $$y=3$$. The value between them, $$x=\frac{-5+17}{2}=6$$ is the location of the vertex. Putting $$x=6$$ in gives the $$y$$-value of the vertex. The picture of the graph falls into place.

To see interesting $$x$$-values in the formula $$y=-3x^{2}+18x-7$$, for example, focus on the “ $$x$$ part” of the equation:

$$y=\underline{-3x^{2}+18x}-7=3x(-x+6)-7$$.

Now we see that $$x=0$$ and $$x=6$$ are interesting for this downward facing U.

COMPLETELY OPTIONAL ABSTRACT-THINKING QUESTION

We have two schema for “seeing” quadratics:

I: As $$y=x^{2}$$ messed around with directly.     As for $$y=3(x-2)^{2}+5$$, for instance.

II: As two interesting $$x$$-values laid bare.    As for $$y=3(x-2)(x-8)+5$$, for instance.

Is schema I already part of schema II in some sense?

What do I mean by this? If we write $$y=3(x-2)^{2}+5$$ as $$y=3(x-2)(x-2)+5$$ then it looks very similar to the example in given for II. There is a “repeated” interesting $$x$$-value.

So … Is there a way to make sense of repeated interesting $$x$$-values so that we never have to think of approach number I and make everything follow from approach number II in some way?

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