## Quadratics

### 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.*

**Answer**:

My first piece of advice:

**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 \(x=3\). □

**YOUR TURN AGAIN …**

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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