### 4.6 More Traditional Textbook Questions

Lesson materials located below the video overview.

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

PRACTICE 87: Here are three quadratics:

(A) $$y=3(x-3)(x+5)$$

(B) $$y=2x^{2}+6x+8$$

(C) $$y=2(x-4)^{2}+7$$

Here are four questions:

i) What is the smallest output the quadratic produces?

ii) Where does the quadratic cross the $$x$$-¬axis?

iii) Where does the quadratic cross the $$y$$-axis?

iv) What are the coordinates of the vertex of the quadratic?

For which of the three quadratics might it be easiest to answer question i)?
For which of the three quadratics might it be easiest to answer question ii)?
For which of the three quadratics might it be easiest to answer question iii)?
For which of the three quadratics might it be easiest to answer question iv) ?

PRACTICE 88:

a) Attempt to solve the equation $$x^{2}+10x+30=0$$. What happens?

b) Sketch the graph of $$y=x^{2}+10x+30$$.

c) Use the graph to explain geometrically why there was is no solution to $$x^{2}+10x+30=0$$.

d) According to the graph, should there be solutions to $$x^{2}+10x+30=11$$.? If so, find them.

e) Find a value $$b$$ so that the equation $$x^{2}+10x+30=b$$ has exactly one solution.

Practice exercise 88 points out a good technique:

IF YOU CAN, DRAW A SKETCH FIRST.
A PICTURE SPEAKS 1001 WORDS!

EXAMPLE 89: How many times does the graph of $$y=-x^{2}+4x-5$$ cross the $$x$$-axis?

Answer: We have $$y=-x(x-4)-5$$ and so $$x=0$$ and $$x=4$$ are interesting. (They both give $$y=-5$$.)

The vertex thus occurs at $$x=2$$ and here $$y=-4+8-5=-1$$.

We see this graph crosses the $$x$$-axis no times. □

PRACTICE 90: How many times does each of the following quadratics cross the $$x$$-axis?

a) $$y=x^{2}-2x-4$$

b) $$y=3x^{2}-12x+12$$

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

How many $$x$$-values give $$y=0$$ for the next two equations?

d) $$y=-500x^{2}-200x+3000$$

e) $$4=4x^{2}+80x+200$$

COMMENT:  You may have been taught a formula “$$b^{2}-4ac$$ ” to answer these sorts of questions. As you can see, we don’t need it!

PRACTICE 91: Find a value $$k$$ so that $$y=-2x^{2}+8x+k$$ has largest value $$43$$.

HINT: Draw a sketch as best as you can, not knowing what $$k$$ should be.

PRACTICE 92: Find a value $$k$$ so that $$y=3(x-10)(x+10)+k$$ just touches the $$x$$-axis.

HINT: Draw a sketch as best you can, not knowing what $$k$$ should be.

EXAMPLE 93: Find a value $$k$$ so that $$y=(x-c)(x+c)+k$$ has smallest value $$-2$$.

Answer: Here $$x=c$$ and $$x=-c$$ are interesting. They each give $$y=k$$.

So we want an upward facing U through these symmetrical points that just reaches down to $$-2$$. The graph must look like this:

By symmetry, the “action” must be happening at $$x=0$$.

Put in $$x=0$$, $$y=-2$$ to see we must have:

$$-2=(-c)(c_+k$$

$$-2-c^{2}+k$$

giving $$k=c^{2}-2$$.  □

PRACTICE 94:

a) Find a value for $$k$$ so that $$y=5x^{2}-10x+k$$ just touches the $$x$$-axis.

b) Find a value for $$m$$ so that $$y=-2x^{2}-18x+m$$ has largest value $$100$$.

c) Find a value for $$a$$ so that $$y=(x-a)(x-3a)$$ has smallest value $$-10$$.

PRACTICE 95: Consider $$y=2(x-1)(x+1)+(x-3)(x+4)+x(x-2)$$.

a) If one were to expand this equation, would one obtain a quadratic expression $$y=ax^{2}+bx+c$$  for some numbers $$a$$, $$b$$ and $$c$$? YES/NO.

b) Explain why we can quickly see that $$a=2+1+1=4$$.

c) Put $$x=0$$ into $$y=2(x-1)(x+1)+(x-3)(x+4)+x(x-2)$$. Explain why we now know $$c=-14$$.

So we have $$2(x-1)(x+1)+(x-3)(x+4)+x(x-2)=4x^{2}+bx-14$$. We don’t know what $$b$$ is yet.

d) What might be an easy way to find the value $$b$$? What is its value?

e) Sketch the graph of $$y=2(x-1)(x+1)+(x-3)(x+4)+x(x-2)$$.

PRACTICE 96: A rectangle has one side of length $$x+3$$ inches and the other of length $$7-x$$ inches. Which value of $$x$$ gives a rectangle of maximal area?

Here’s a question that is designed to give the impression that quadratics are used in the “real world.” The trouble is that companies rarely know a formula for their revenues, and if they did, it is very unlikely that the formula would be a perfect quadratic equation.

I would guess these questions are designed more to test if you can think about formulas and apply meaning to their properties: What does the vertex mean? What does an intercept mean? And so on.

Just to acknowledge the silliness of the “real world” meaning of these questions, I’ve designed a question here about floogle.

EXAMPLE 97: A company produces and sells floogle by the pound. Its revenue $$R$$ and cost $$C$$ functions are given by:

$$R(x)=-2x(x-600)$$

$$C(x)=200x+80,000$$

where $$x$$ represents the number of pounds of floogle produced per day and $$R$$ and $$C$$ are each in units of dollars.

a) Give rough sketches of these two functions on the same set of axes.

b) What is the “break even” point for the company?

c) For which value of $$x$$ is the revenue $$R$$ at a maximum?

d) For which value of $$x$$ is the profit the company makes at a maximum? What is that maximum profit?

Answer: Firstly we should make clear as to what “revenue” and “cost” mean. Revenue usually means “how much money is brought in.” So $$R(x)$$ is the number of dollars brought in by making and selling $$x$$ pounds of floogle. Of course there are costs involved in producing floogle, and $$C(x)$$ is the number of dollars it costs to make $$x$$ pounds of floogle.

Apparently, $$R(x)$$ is given by a downward facing parabola. I guess this means that if you start making too much floogle you flood the market and can’t sell as much. The revenue starts going down after some point.

The cost function $$C(x)$$ is a straight-line formula: $$C(x)=200x+80000$$. Before you make any floogle, you have a cost of $80,000. (The cost of the factory?) Thereafter it costs an additional$200 for every pound of floogle you make. (Expensive stuff!)

Okay … On to the mathematics. As part a) suggests, pictures are always good.

a) $$R(x)=-2x(x-600)$$ has $$x=0$$ and $$x=600$$ as interesting, each giving an output of zero. The vertex must occur at $$x=300$$ and here $$R(300)=-2\cdot 300(-300)=180,000$$.

$$C(x)=200x+80,000$$. This has a straight-line graph of slope $$200$$ starting on the $$y$$-axis at position $$80,000$$. Just to be clear, at $$x=300$$, we have $$C(300)=200 \cdot 300 + 80000=140,000$$, which is lower than the vertex of the quadratic.

The graphs look like:

b) “Break even” must be the point when the revenue starts to outweigh the costs. This is a very important thing for companies to know: How many pounds must they sell in order to start making a profit. We see from the picture that there is a point after which $$R(x)$$ starts to be larger than $$C(x)$$.

To find this point we must solve:

$$R(x)=C(x)$$

$$-2x(x-600)=200x+80,000$$

$$-2x^{2}+1200x=200x+80,000$$

$$-2x^{2}+1000x=80,000$$

Dividing through by $$-2$$ gives:

$$x^{2}-500x=-40,000$$

The box method (check this!) suggests to add $$62,500$$ to each side:

$$x^{2}-500x+62500=22500$$

$$(x-250)^{2}=22500$$

$$x-250=150$$  or  $$x-250=-150$$

$$x=400$$  or $$x=100$$

From the graph we expect two answer to this, but it is the left answer, $$x=100$$, that is the break even point.

c) The revenue is at a maximum for $$x=300$$.

d) Now here is a surprise for students: The profit is not a maximum at $$x=300$$!

The amount of profit $$P$$ the company makes is given by revenue minus cost:

$$P(x)=R(x)-C(x)$$

and this is its own formula:

$$P(x)=-2x(x-600)-(200x-80000)$$

$$=-2x(x-600)-200x-80000$$

$$=-2x^{2}+1200x-200x-80000$$

$$-2x^{2}+1000x-80000$$

It is a downward facing U with two interesting $$x$$-values:

$$P(x)=-2x(x-500)-80000$$.

Its vertex occurs halfway between the two interesting $$x$$-values, at $$x=250$$.

The maximum profit is:

$$P(250)=-2 \cdot 250(-250)-80000=45,000$$.

PRACTICE 98: OPTIONAL THINKING QUESTION

In part c) we found that the $$R(x)$$ and $$C(x)$$ curves cross at $$x=100$$ and $$x=400$$, and in part d) we found that the maximum profit occurs at $$x=250$$. Notice that $$250$$ is halfway between $$100$$ and $$400$$. Is this a coincidence?

PRACTICE 99: Repeat question 97 but for a company that produces and sells gloop by the pound. Their revenue and cost functions, in dollars, are given by:

$$R(x)=240x-3x^{2}$$

$$C(x)=1800+30x$$

where $$x$$ is the number of pounds of gloop made and sold.

PRACTICE 100: A THEORY QUESTION

Consider the quadratic ](y=ax^{2}+bx+c\). Rewrite this as:

$$y=x(ax+b)+c$$.

a) What are the two interesting $$x$$-values for this quadratic?

b) Explain why the vertex of its graph occurs at $$x=-\frac{b}{2a}$$.

COMMENT: Many curricula have students memorise this result. For example, given $$y=3x^{2}+4x+8$$, say, they like students to be able to say that its vertex lies at $$x=-\frac{b}{2a} = -\frac{4}{2 \cdot 3} = -\frac{2}{3}$$. If speed is important to you, then great! If not, there is nothing wrong with writing $$y=x(3x+4)+8$$ and saying that the vertex is halfway between $$x=0$$ and $$x=-\frac{4}{3}$$.

PRACTICE 101: Here’s to the power of interesting -values! Consider this challenge.

Sketch a graph of $$y=(x+10)^{3}(x+6)^{2}(x+2)(x-3)^{4}(x-5)(x-12)^{37}$$.

We have not done this before so we are going to have to use our common sense.

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

Where does the graph cross the $$x$$-axis?

When $$x$$ is a huge positive number does the graph want to be huge and positive?

What does the graph want to be if $$x$$ is a huge negative number like $$-1000000$$?

Between the places where the graph crosses the $$x$$-axis, can you tell if the graph wants to be positive or negative?

If you can answer these questions then you have enough information to make a pretty good sketch of the thing!

See the COMPANION GUIDE to this QUADRATICS course for an essay on how to graph the imaginary solutions to a quadratic with no real solutions! (Cool!)

## Self Check: Question 16

Another self-check!

## Self Check: Question 17

A second question … jolly ho!

## Self Check: Question 18

One final question:

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