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:

Q4_5_pic1

 

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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