Quadratics

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Foundations of Quadratics

Discussion

A farmer plans to build a rectangular pen with one side cut off by a river, as shown below. He has 40 meters of fence to enclose the remaining three sides.

sheep-pen

Write a function to describe the area A enclosed by the fence, in terms of only the width x.

We know that the perimeter is equal to two times the width (x) added to the height (h). Since we have 40 m of fence total, this relationship can be modeled by the equation

2x+h=40

Since we want an expression entirely in terms of x, we should find a way to represent h in terms of x:

h=40−2x

We know that the area enclosed by the fence is equal to the width multiplied by the height:

A=x×h

hence, we can substitute in:

A=x(40−2x)=40x−2x2

Checkpoint

If x=7m, what is A?

Select the correct option

Even armed with the equation A=−2x2+40x, the farmer still feels lost when trying to understand what this really means about the relationship between x and A. He decides to plot some values to get a more clear picture.

After plugging in some values for x into the equation directly, the farmer obtains the following table:

x	A
2	−2(2)2+40(2)=72
6	−2(6)2+40(6)=168
10	−2(10)2+40(10)=200
14	−2(14)2+40(14)=168
18	−2(18)2+40(18)=72

Discussion

The farmer realizes there are many points in between each x value listed in the table. He decides to plot the values of A that correspond to all of the x values strictly between 0 and 20.

What do you notice about the figure created?

The graph is a smooth, continuous curve with a single highest point.

That point lies on a vertical line of symmetry: if you shift the same distance left or right from its x-coordinate, you land on twin points with identical y-values.

Discussion

What do you notice about the graph above?

The key idea is that the height of each column is corresponds directly to the square of it's x-value.

This leads us to two key observations about parabolas:

Symmetry: When we square a number, the sign of x gets eliminated. So the column height at distance x and −x will always be the same, relative to the middle of the figure.
Steepness of curve: The further away we move from the center, the more squares get added. The number of squares goes as 1,4,9,… Notice how we increase by 1 square, then gain an additional 3, then 5, and so on so forth.

Vertex and Axis of Symmetry

The graph of a quadratic function has the general shape of a parabola.

It is symmetrical about the axis of symmetry and has a maxima or minima at the vertex, which lies on the axis of symmetry.

Checkpoint

Examine the following graph and find:

The coordinates of the vertex.
The equation for the axis of symmetry.

Select the correct option

General form of a quadratic

A quadratic in x is an expression of the form

ax2+bx+c

where a=0.

Checkpoint

We are given the following quadratic:

y=7x2−5x+2

State the value of…

Select the correct option

Discussion

Take a look at the graph below of y=ax2, which has a vertex at (0,0). Notice that adjusting the value of a affects the "width" and the direction of the parabola.

What patterns do you observe? For which values of a does the parabola open upwards? Downwards? When does it appear wider? Narrower?

As you change the value of a in the equation

y=ax2

you should observe the following patterns:

Direction of opening:
- If a>0, the parabola opens upwards.
- If a<0, the parabola opens downwards.
- If a=0, y=0 for all values of x.
Width of the parabola:
- If ∣a∣>1, the parabola is taller (and therefore narrower) than y=x2.
- If 0<∣a∣<1, the parabola is shorter (and therefore wider) than y=x2.

Concavity of a parabola

The concavity of a parabola describes whether it "opens" up or down.

The parabola corresponding to ax2+bx+c is:

Concave up if a>0
Concave down if a<0.

Discussion

Now suppose the farmer wants to enclose exactly 128 m2 with the fence.

Derive an equation in terms of x that allows the farmer to enclose the desired area using the entire length of fencing available.
Find the value(s) of x that satisfy this equation.

Part (a)

We already know how to find the area A depending on the width x:

A=40x−2x2

Enclosing exactly 128 m2 means

40x−2x2=128

Part (b)

We know that there must be two values of x that satisfy the equation. The graph of A=128 intersects with the graph A=−2x2+40x at two distinct points:

To find these points of intersection, we must solve the equation:

40x−2x2=128

for x. To start off, it's extremely helpful to move all the terms over to one side:

−2x2+40x−128=0

Additionally, we should factor out a two in order to reduce the coefficients as much as possible:

x2−20x+64=0

Our next goal is to write the left side as a product. Why?

Recall the zero product property: if (something)(something else) = 0 then at least one of the "somethings" must be equal to zero—which means we have an x value that solves the equation.

For any numbers α and β, consider the general equation (x−α)(x−β)=0.

This equation leads us to some key observations:

If x=α or x=β, the whole left side equals zero, and we have a solution to the equation.
(x−α)(x−β)=0 expands to x2−(α+β)x+αβ=0, which has the same form as our equation, x2−20x+64=0.

With this in mind, it makes sense to equate our equation above to an equation in the form x2−(α+β)x+αβ=0.

x2−20x+64=x2−(α+β)x+αβ

For these to match we need:

α+β=20, αβ=64

Intuitively, we can quickly tell both numbers must be positive since their sum and product are positive. We check the factors of 64:

64⋅1→ sum =65
32⋅2→ sum =34
16⋅4→ sum =20. Aha!

So we can set α=16 and β=4 (or the other way around):

x2−20x+64=(x−α)(x−β)=(x−4)(x−16)=0.

So if x=16 or x=4, we satisfy the farmer's equation since the entire left side becomes zero. So each of these values of x guarantees an area of 128 m2 while maintaining a perimeter of 40 m.

Note that if we wanted to obtain the full factored expression (instead of just solving for x directly), we have to remember the −2 we factored out earlier:

−2x2+40x−128=−2(x−16)(x−4)

Factored form of quadratic

Most quadratics can be factored as a product of linear terms:

a(x−α)⋅(x−β)

We call the generalized form above factored form. Notice that α and β are the roots of the quadratic, since when x=α or x=β the expression will evaluate to zero.

Checkpoint

Factor x2+9x+20.

Select the correct option

One important step to remember: whenever every term shares a common factor, divide it out first:

−5x2+15x−20=0⇒−5(x2−3x+4)=0⇒x2−3x+4=0

Factoring is easiest when the x2 term has a coefficient of 1—you just look for two numbers that multiply to c and add to b. Often you can reach this “nice” setup by dividing out a common factor first.

But if the coefficient in front of x2 isn’t 1, factoring gets a bit trickier:

Factoring by Inspection

We can factor quadratics in the form ax2+bx+c by splitting b into a sum α+β such that that αβ multiplies to ac.

After rewriting the expression as ax2+αx+βx+c, factor the first pair and the second pair separately. The common factor will emerge, and you can pull it out. The result is the fully-factored expression.

For example, in the quadratic 3x2+8x−3, we want to split 8 into α+β such that αβ=−9. We can do this by choosing α=9 and β=−1:

3x2+αx+βx−3=3x(x+3)−1(x+3)=3x2+9x−x−3=(3x−1)⋅(x+3)

Exercise

Factor 2x2+19x+35 completely.

Select the correct option

Finding the roots also helps us interpret the quadratic graphically:

Quadratic x-intercepts

The roots of a quadratic correspond to the x-intercepts of its graph. When x=a or x=β, the entire expression equals zero, which is reflected on the graph.

The equation of the parabola below is −(x−α)(x−β):

Discussion

Factor the expression A=−2x2+40x to find the x values that make the area zero.

Now let's suppose the farmer keeps the same 40 m of fence but now wants the largest possible area.

What value of x would achieve this?
What would be the maximum resulting area?

Part (a)

We set the area function to zero and factor:

A=−2x2+40x=0

Factor out the common term −2x:

A=−2x(x−20)

Solving −2x(x−20)=0 gives

x=0orx−20=0⟹x=20

Hence the area is zero when x=0 or x=20.

Part (b)

Let's look back at the graph of A:

We're looking for the value where the question mark is. As we've already discussed, parabolas are symmetric about their axis of symmetry.

We can see from the graph that the vertex lies on this axis of symmetry. Because of the symmetry, it therefore lies halfway between the two roots that we found:

xmax=21(0+20)=10

Part (c)

At this point, we know that the x-coordinate that maximizes the area is 10.

To find the corresponding area A, we can substitute directly into the equation A=−2(10)2+40(10)=200.

So the maximum area enclosed by the fence is 200 square meters.

Vertex height form

Recall our earlier "square stack" graph:

This helped us understand the behavior of y=x2, where the height is the square of the horizontal distance from the vertex. That sounds like a jumble, but all it means is height =(x units from vertex)2.

But now, imagine we slide it's lowest point from (0,0) to the coordinates of a new vertex (h,k).

Take a look at the graph below, and move around the vertex by dragging it. What do you notice? How does moving the vertex affect our "square stack" relationship?

As you adjust the vertex, notice that the entire square construction moves with it. The whole quadratic has the same shape and form relative to the shifted vertex.

This leads us to a key question: how can we mathematically express the quadratic in terms of the vertex coordinates?

First, notice that the number of squares in each stack still depends on their horizontal distance from the vertex, which is now located at x= h.

The distance from the vertex is equal to ∣x−h∣. Remember that absolute value (∣⋯∣) ensures we measure the horizontal distance regardless of whether the x value in question is to the left or right of h.

Since we know that height = (x units from vertex)2, we can substitute in directly:

height =(∣x−h∣)2=(x−h)2

Note that squaring already guarantees the height is positive, so we can remove the absolute value. This formula for height accounts for any horizontal changes in the vertex, while maintaining the shape and form we saw graphically.

Additionally, when we vertically shift the vertex from 0 to k, we're just raising (or lowering) every point by the same amount, i.e.

total height (y) =(x−h)2+k

Finally, there's one more element to consider: the scale factor a. Take a look at the graph below and notice how adjusting value of a affects the "square stack":

Notice that the vertex still remains in the same place as we scale the parabola by a, but the squares are "stretched" depending on the value of a.

There are two things to notice:

The squares are being stretched into rectangles, but the number of shapes in each column is not changing, it's still (x−h)2. The height of each column is thus a(x−h)2.
The vertex does not move - it's height above the x-axis is still k.

All of this comes together to create our finished form:

Vertex Form & Coordinates

A quadratic in the form ax2+bx+c can be written in the form

a(x−h)2+k

for some h,k∈R.

Notice that the minimum / maximum (depending on a) of this expression occurs when (x−h)2=0, ie x=h.

The resulting y-coordinate is then simply k. Thus the coordinates of the vertex are (h,k). We call this form the vertex height form as it illustrates the connection between the quadratic and its vertex.

Exercise

Find an equation for the parabola with vertex (4,1) passing through the point (2,9).

Select the correct option

At this point, it's clear that we can express every parabola ax2+bx+c in vertex form: a(x−h)2+k

But how can we convert from the general form to vertex form algebraically?

Let's expand vertex form fully:

a(x−h)2+k=ax2−2ahx+ah2+k

and set it equal to ax2+bx+c, to see how the coefficients match up:

ax2−2ahx+ah2+k=ax2+bx+c

So the ax2 already lines up, and we can see that

⎩⎪⎨⎪⎧−2ahx=bx⟹−2ah=b⟹h=−2ab ah2+k=c

Recall that the axis of symmetry has the equation x=h (the x-coordinate of the vertex). This reveals how the axis of symmetry can be found directly:

Equation of the axis of symmetry

For the quadratic ax2+bx+c, the parabola has an axis of symmetry at

x=−2ab📖

The axis of symmetry is the vertical line dividing the parabola perfectly in 2. The x-coordinate of the vertex, h, is equal to the x value where the axis of symmetry is located.

Exercise

State the vertex of 4x2−8x+3.

Select the correct option

Completing the square

Suppose we want to convert the parabola 3x2+12x−18 into vertex form.

First, identify the coefficients. We can see that a=3, b=12, and c=−18. The axis of symmetry is located at

x=−2ab=−2(3)12=−2.

In vertex form, the axis of symmetry is located at the x-coordinate of the vertex (h,k). So h=−2.

Filling out the general vertex form with what we know:

a(x−h)2+k3(x−(−2))2+k=3(x+2)2+k

Expanding fully:

3(x+2)2+k=3x2+12x+12+k

We're so close! All that's left to do is to solve for k by matching the constant terms (eg. terms without x):

12+k=c12+k=−18k=−30

Completing the square

To convert from the form ax2+bx+c to a(x−h)2+k:

The values for a will match up directly.
Use the axis of symmetry x=−2ab to find h=−2ab.
Equate c=ah2+k, and substitute the h found to find k.

At this point, we have identified the three forms that are used to describe quadratic expressions. Each of them has it's own benefits:

Form	Prototype	What you can read off instantly
Standard (General)	ax2+bx+c	a: opens up/down & stretch y-intercept at (0,c)
Factored	a(x−p)(x−q)	a: same meaning Roots (x-intercepts) at (p,0) and (q,0)
Vertex	a(x−h)2+k	a: same meaning Vertex located at (h,k) Axis of symmetry: x=h

Here are some examples of equivalent quadratic expressions in all three forms:

Standard form	Factored form	Vertex form
2x2−8x+6 Concavity: opens up (a=2>0) y-intercept at (0,6)	2(x−1)(x−3) Concavity: opens up (a=2>0) Roots at x=1,3	2(x−2)2−2 Concavity: opens up (a=2>0) Vertex at (2,−2)
−3x2+12x−9 Concavity: opens down (a=−3<0) y-intercept at (0,−9)	−3(x−1)(x−3) Concavity: opens down (a=−3<0) Roots at x=1,3	−3(x−2)2+3 Concavity: opens down (a=−3<0) Vertex at (2,3)
4x2+4x−15 Concavity: opens up (a=4>0) y-intercept at (0,−15)	4(x+25)(x−23) Concavity: opens up (a=4>0) Roots at x=−25,23	4(x+21)2−16 Concavity: opens up (a=4>0) Vertex at (−21,−16)

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Some quadratics don't factor nicely with integers. In this case, we can use the quadratic formula, a universal tool for solving a quadratic.

Derivation of the Quadratic Formula

(optional)

Let's try to set up ax2+bx+c=0 in terms of x directly, since this would be a general solution to solve for the roots of any quadratic.

Let's start by recalling the axis of symmetry x=−2ab, is the x coordinate of the vertex. The quadratic can thus be rewritten:

a(x+2ab)2+k=0⇒a(x+2ab)2=−k

What is the value of k? The constant terms in vertex form must match the constant term c:

a⋅4a2b2+k=c⟹k=c−4ab2

Therefore

a(x+2ab)2=4ab2−c⟹x+2ab=±√4a2b2−ac

Multiplying both sides by 2a:

2ax+b=±√b2−4ac⟹2ax=b±√b2−4ac

Finally, we isolate x on the left side of the equation to obtain the (iconic!) quadratic formula:

x=2a−b±√b2−4ac

This completes the derivation. Note that you are not required to memorize this proof, but should feel comfortable substituting in values for a,b and c to solve for the roots.

Quadratic formula

For any quadratic ax2+bx+c, the roots of the quadratic can be found using the quadratic formula:

x=2a−b±√b2−4ac📖

Exercise

Find the roots of 5x2+6x−1=0 using the quadratic formula.

Select the correct option

Sometimes, solving for the roots is very difficult:

Solving Quadratics with a Calculator

When the values of a,b or c are large enough that using the quadratic formula becomes difficult, a calculator can be used to find the roots.

Your calculator should have an app for solving quadratics.

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