Quadratic EquationsGraphing Quadratics

At the beginning of this chapter, we already saw what happens if we plot the function y=x2 in a coordinate system.

The function y=x^2 is a symmetrical, U-shaped graph.

NOTE: Is “U-shape” an accurate term? The sides of the letter U look vertical. Do the sides of y=x2 ever become perfectly vertical? If we accurately plot the graph of y=x2, to what extent is it actually U like?

The place where a U-shaped graph dips down to its lowest point, if it is upward facing, or “dips up” to its highest point if it is downward facing is called the vertex of the quadratic graph.

Shifting and Sliding

NEED Example for why to shift parabola.

I am six feet tall and am standing at the position x=4 feet on the horizontal axis. Is it possible to write down a formula for a function whose graph is the same U-shaped curve as for y=x2 but positioned to balance on my head?

Perhaps try this before reading on. Play with some possible formulae. Plug in some sample points and table values. Test whether your ideas offer any hints as to a solution to this challenge.

Shifting Vertically

How would the graph of y=x2+3 appear?

Notice that this new function is adds three units to each output:

This has the effect of raising the entire graphs three units in the vertical direction:

The graph of the function y=x2−5 would be the same graph shifted downwards 5 units, and the graph of y=x2+3–√ would the graph shifted upwards 3–√ units.

PRACTICE 58:

a) Sketch y=x2−5 b) Sketch y=x2+1/2.

Shifting Horizontally

One thing to notice is that this graph has a “dip” at x=0.

Now consider the function y=(x−3)^2.

Notice that when we put x=3 into this formula we obtain the output 02. That is, the number 3 is “behaving” just like x=0 was for the original function.

In y=(x−3)^2 we have that 3 is the “new zero” for the x-values.

So whatever the original function was doing at x=0, it is now doing it at x=3.

The original function y=x2 dips at x=0, so the graph of the function y=(x−3)^2 dips at x=3.

The entire graph has been shifted horizontally – and one can check this by drawing a table of values if you like:

Slider move

PRACTICE 57:

a) Sketch a graph of y=(x−4)2.

b) Sketch a graph of y=(x+12)2.

Both Together

Here’s the graph of y=x2:

and here is the graph of y=(x−2)2. Here “2” is acting as the new zero for the x-values, so the dip that was occurring at zero is now occurring at 2:

Now consider y=(x−2)2+3 . This is the previous graph shifted upwards three units:

Note that the order doesn't matter.

PRACTICE 59:

a) Sketch y=(x−4)2−1.

b) Sketch y=(x+4)2+2.

c) Sketch y=(x+1)2−2.

PRACTICE 60: I am six feet tall and am standing at the position x=4 feet on the horizontal axis. Write down the formula of a U-shaped graph that sits balanced on my head.

Scaling and Squashing

What can we say about the graph of y=2x2 ? Certainly all the outputs are doubled:

This creates a “steeper” U-shaped graph:

PRACTICE 61:

a) Draw a table of values for y=3x2 and sketch its graph.

b) Which graph is “steeper,” that for y=100x2 or that for y=200x2?

Consider y=−x2. It’s outputs are the negative of the outputs for y=x2:

PRACTICE 62: Describe the graph of y=−2x2.

PRACTICE 63:

a) Quentin says that the graph of y=110x2 will be a very “broad” upward facing U-shaped graph. Is he right? Explain.

b) Describe the graph of y=−11000x2.

Putting it all together:

(1) Which of the following could be a formula for the following graph?

y=4/3(x−1)^2+1 y=4/3(x+1)^2+1 y=4/3(x−1)^2−1 y=4/3(x+1)^2−1 y=−4/3(x−1)^2+1 y=−4/3(x+1)^2+1 y=−4/3(x−1)^2−1 y=−4/3(x−1)^2−1

(2) A quadratic graph is symmetrical about the vertical line x=3 and has highest value −17. Which of the following could be a formula for that quadratic?

y=200(x−3)^2+17 y=−200(x−3)^2+17 y=200(x−3)^2−17 y=−200(x−3)^2−17

(3) If y=a(x+b)2+c is a quadratic passing through the origin with vertex (2,3), then a+b+c equals …

1/4 1 3/4 4 1/4 5 1/4

(4) Write down the formula for a quadratic which crosses the x-axis at x=−10 and at x=−6 and crosses the y-axis at y=−20. Sketch this in your mind’s eye: Will this quadratic be upward-facing or downward-facing?

(5) Will this very steep, upward-facing quadratic y=10,000,000x2+40,000,000 ever cross the vertical line x=999,999,999,999,999?

YES NO

Normal Form and Vertex Form

Here is the graph of y=x2:

Here is the graph of the line y=x:

What do we obtain if we “add” these two formulas together and graph y=x2+x?

To the right, it is clear that “ x2 plus x” adds together two positive numbers: The graph of y=x2+x is clearly getting larger and larger as we go over to the right.

Matters are less clear to the far left: The x2 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=x2+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. 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=−12 we have y=(−12)2+(−12)=−14=−0/25.

At x=−34 we have y=916−34=−316=−0.1875.

At x=−14 we have y=116−14=−316=−0/1875 The graph seems to dip down to lowest value −14 at x=−12.

Is this a perfect U-shaped graph, with x=−12 behaving as zero for the x-values and shift −14 units? That is, could we have y=a(x+12)2−14 for some steepness a?

ACTUALLY … If y=a(x+12)2−14 goes through the point (0,0) then we need 0=14a−14 giving a=1.

So is y=x2+x really the U-shaped graph y=a(x+12)2−14 in disguise?

To answer that question let’s expand (x+12)2:

(x+12)2=x2+12x+12x+14=x2+x+14

and so

(x+12)2−14=x2+x+14=14=x2+x.

YES! y=x2+x is a perfect U-shaped quadratic graph with vertex at (−12,−14).

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

Every quadratic y=ax2+bx+c is a version of y=x2 “messed around with”.

It equals a(x+something)2+something with steepness a.

As such, the graph of y=ax2+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!

The Power of Symmetry

It is still hard to determine where the vertex of this graph lies, and so it is still hard to actually graph the curve.

If we can find two interesting and symmetrical x-values, then common sense will point the way! Let me explain what I mean by this.

EXAMPLE 76: Sketch y=2(x−3)(x−9)+7.

Answer: The first thing to note is that if we were to expand 2(x−3)(x−9)+7 we would obtain an expression of the form ax2+bx+c . (Actually, check this: Show that2(x−3)(x−9)+7 expanded equals 2x2−24x+34.)

Without doing all the expanding, however, we can see that the positive 2 in front will help give the term 2x2 in the expansion. We have a positive steepness.

So we know y=2(x−3)(x−9)+7 is going to give a symmetrical upward-facing U-shaped graph.

In y=2(x−3)(x−9)+7 two obvious and interesting x-values are staring us in the face:

Put x=3 and we get y=0+7=7. Put x=9 and we get y=0+7=7.

So we now have an upward-facing U that goes through two symmetrical points:

Since the graph is symmetrical, common sense says that the graph must have vertex at x=6, the point halfway between x=3 and x=9.

The trouble is we don’t know how high the U is at this point: Does the U sit above the -axis? Dip below it? Just touch it?

Well … let’s just plug x=6 into the formula to find out!

When x=6 we have y=2(3)(−3)+7=−11.

We must have the following picture:

DONE! □

NOTICE: I am personally not at all fussed about scale in the picture nor extra-fussy details about the graph. As long as the information given on the diagram is clear and correct and gives all the key features of interest, then all is good.

HOW DOES ONE FIND INTERESTING x-VALUES IF THEY ARE NOT OBVIOUS?

Consider y=x2+4x+5. This is going to be an upward-facing U-shaped graph.

To find interesting x-values, focus on the “x part” of the formula, in this case, just: x2+4x.

The only thing I can think to do with is to rewrite it as x(x+4) and so:

y=x(x+4)+5.

Now it is clear that x=0 and x=−4 are interesting. They both give y=5.

By symmetry, we know the vertex lies midway between these interesting x-values, at x=−2, and here y=(−2)(2)+5=1.

Here then is its graph.

PRACTICE 81: Make reasonable sketches of the following quadratics:

a) y−x2−2x+3 b) y=4x2+8x+5 c) y=−5x2−20x+10 d) y=−2(x+30)(x−50)

PRACTICE 82: Make reasonable sketches of the following quadratics:

a) y=3(x−1)2+5 b) y=(x−2)(x+3)+8 c) y=−5x2−20x+10 d) y=7(x−2)(x+3)

PRACTICE 83: A quadratic crosses the x-axis at x=2 and at x=8, and it crosses the y-axis at y=7. Find a formula for the quadratic.

HINT: Make a sketch of the information you have so far and use your common sense to see what the quadratic must be doing.

PRACTICE 84: Find a formula for the following quadratic:

Old Content

To better understand the relationship between quadratic equations and their corresponding graphs, we have to think about transformations. Any

All parabolas can be thought of as some transformation of the standard parabola which has the equation $y=x^2$y=x2.

If we compare $y=x^2\left(0,0\right)$(0,0)) to $y=a\left(x-h\right)^2+k$y=a(x−h)2+k, then we can think of the second equation and its graph as a transformation of $y=x^2$y=x2:

$h$h represents the number of units that $y=x^2$y=x2 is translated horizontally, and $k$k represents the number of units that $y=x^2$y=x2 is translated vertically. Note that with the direction of the translation:

if the equation is $y=\left(x+1\right)^2-3$y=(x+1)2−3, the horizontal translation is 1 unit to the left. if the equation is $y=\left(x-1\right)^2-3$y=(x−1)2−3, the horizontal translation is 1 unit to the right.

These movements are called transformations. Transform means change, and these transformations change the simple quadratic $y=x^2$y=x2 into other quadratics by moving (translations), flipping (reflecting) and making the graph appear more or less steep (dilating).

$y=a\left(x-h\right)^2+k$y=a(x−h)2+k Is called turning point form.

it tells us the turning point immediately, and knowing the turning point we can draw a pretty good sketch of any quadratic it explains to us a number of transformations that have occurred to the quadratic from the simple quadratic $y=x^2$y=x2. Identifying transformations

The form $y=a\left(x-h\right)^2+k$y=a(x−h)2+k shows us:

This lesson focuses on the connection between the equations of quadratics and their graphs, and how we can use the idea of transformations to more easily identify and make sense of these quadratics.

The concavity of a parabola is determined by the $a$a value in the equation $y=ax^2$y=ax2.

$y=-x^2$y=−x2 vs $y=x^2$y=x2 : The sign of the coefficient $a$a determines whether the parabola is concave up or concave down.

$y=a\left(x-h\right)^2+k$y=a(x−h)2+k

$a$a determines concavity

$k$k represents the vertical shift upwards or downwards (look at its sign to determine the direction)

$h$h represents the horizontal shift to the right or to the left (look at its sign to determine the direction)

Given a quadratic function and it's equation, there are many different questions we could ask about their properties:

• Does the parabola point up or down?
• What are the coordinates of the turning point?
• Where does the parabola cross the x-axis (the Zeros)?
• Where does the parabola cross the y-axis?
• How stretched or squashed is the parabola?

Sketching a parabola from the equation

x-coordinate-sketch(validate="quadratic:2,-4,1")

Extracting the equation from a sketch

Graphing the quadratic Let's look at the quadratic $y=-2\left(x-3\right)^2-3$y=−2(x−3)2−3

Shape - we can see that the quadratic will be concave down because which is < $0$0

Vertex - we can see that the quadratic will have turning point at (3,−3).

With just these two pieces of information we can roughly sketch the curve.

But many parabolas are concave down and have a vertex at (3,−3). To be able to graph the parabola more accurately with the correct amount of steepness, we need another piece of information. We need another point on the graph.

A common point to use would be the $y$y intercept.

Remember intercept occurs where the value of $y$y is $0$0 intercept occurs where the value of $x$x is $0$0

Now we can sketch the curve.

Using vertex , the concave down shape and a $y-21$ :

As a hand drawn sketch, this shows all the information we need to sketch the quadratic.

Features of Quadratic Functions

Remember the skateboards we built at the beginning of this course? Some of the first prototypes are ready, so let's build a small skate park to try them out! We’ve got 50m of fence, as well as two existing walls of the factory building we can use to enclose the area.

interactive diagram

Using mathematics we can work out what x has to be, so that the area of the skate park is as large as possible.

If we call the horizontal part of the fence x, then the vertical part has length [{x-equation.var(vars="x", fns="+ -")}]. The total area of the skate park is

A=[{x-equation.var(vars="x", fns="+ -")}]

Once again, we have a quadratic equation! Except, rather than solving it, we want to find it’s maximum.

Definition of vertex and turning point

Diagram, numeric solution

Scheitelpunktform und Grundform

There are two good ways to think about the graph of a quadratic equation y=ax2+bx+c:

• I: As y=x2 messed around with. 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 (−10,17).)

• II: As two interesting x-values laid bare. 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=−5+172=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.

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?

Questions

EXAMPLE 85: Here is a traditional algebra II question:

Consider y=2x2−8x+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 (2,−2).

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=2x2−8x+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=2x2−8x+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. □

PRACTICE 86: Consider y=5x2−10x−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?

PRACTICE 87: Here are three quadratics:

(A) y=3x−3x+5 (B) y=2x2+6x+8 (C) y=2x−42+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 each question?

PRACTICE 88:

a) Attempt to solve the equation x2+10x+30=0. What happens?

b) Sketch the graph of y=x2+10x+30.

c) Use the graph to explain geometrically why there was is no solution to x2+10x+30=0.

d) According to the graph, should there be solutions to x2+10x+30=11? If so, find them.

e) Find a value b so that the equation x2+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=−x2+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=x2−2x−4 b) y=3x2−12x+12 c) y=2−x2

How many x-values give y=0 for the next two equations? d) y=−500x2−200x+3000 e) 4=4x2+80x+200

COMMENT: You may have been taught a formula “b2−4ac ” to answer these sorts of questions. As you can see, we don’t need it!

PRACTICE 91: Find a value k so that y=−2x2+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−c2+k giving k=c2−2. □

PRACTICE 94:

a) Find a value for k so that y=5x2−10x+k just touches the x-axis.

b) Find a value for m so that y=−2x2−18x+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=ax2+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)=4x2+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?

PRACTICE 100: A THEORY QUESTION

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

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

b) Explain why the vertex of its graph occurs at x=−b2a.

COMMENT: Many curricula have students memorise this result. For example, given y=3x2+4x+8, say, they like students to be able to say that its vertex lies at x=−b2a=−42·3=−23. 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=−43.

PRACTICE 101: Here’s to the power of interesting x-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!

QUESTION: i) What is the x-coordinate of the vertex of y=−4x2+10x+7? ii) What input x gives the smallest output for y=2x2−8x−100? iii) What is the “line of symmetry” of y=(x−4)2+5?

QUESTION: Let’s pretend that a company that makes and sells mattresses has identified formulas for its revenue R(x) brought in from selling and cost C(x) in making x mattresses per week. Those formulas are:

R(x)=100+20x−x2 C(x)=2x+145

• Sketch graphs of these functions on the same set of axes.
• What number of mattresses should they make and sell per week in order to break even?
• What number of mattresses should they make and sell per week in order to maximize their profit?

QUESTION: Find a formula for k in terms of c so that the quadratic y=x(x−2c)−k just touches the x-axis.