5.2 Quadratic Functions

Learning Objectives

By the end of this section, you will be able to:

• identify quadratic functions and their important features: vertex, roots, end behavior
• write quadratic functions in standard and vertex forms
• assess the domain and range of a quadratic function
• find real roots of quadratic functions using factoring techniques
• sketch graphs of quadratic functions

Next up in our tour of polynomial functions, you will see the degree-two polys coming up here on my left, your right. These are known as quadratic functions. For this section, you may want to review Section 2.2 Factoring Techniques if you haven't already, because I am going to assume that you're generally proficient with those methods. You will have to solve quadratics ad nauseum in Calc I optimization problems, so make sure you come out of this section with solid mastery!

Quadratic Functions

A quadratic function is a degree-two polynomial function, i.e. it can be written in the general or standard form

\[ f(x) = a_2 x^2 + a_1 x + a_0 \quad \text{ or more often } \quad f(x) = ax^2 + bx + c \notag \]

where \(a(\neq 0), b,\) and \(c\) are constants. Fun facts about quadratic functions:

• The graphs of these functions are parabolas, which can open upward or downward. If the leading coefficient \(a > 0\), it opens upward. If \(a < 0\), it opens downward. There is a point called the vertex, which is the peak of the hill if the parabola opens downward, or the bottom of the trough if the parabola opens upward. The parabola is symmetric about the vertical line passing through the vertex.
• Quadratics are often written in vertex form, \( f(x) = a(x-h)^2 + k\), highlighting the location of the parabola's vertex, \( (h,k)\). ALERT: be careful with the signs because of that subtraction!
• The domain of a quadratic function (unless specified due to an application) is all real numbers, \(\mathbb{R}\).
• The range of a quadratic function is limited due to its parabola shape. Identifying the vertex will help you determine the range, since it will give the maximum or minimum possible function value.
• The \(x\)-intercept(s) of quadratic functions are called roots or zeros. As always, to find these, set the whole function equal to 0 and solve for \(x\).

Let's look at some graphs while we're here talking about parabolas. Quadratic functions look something like this:

Depending on whether the parabola opens upward or downward, the quadratic function will either have a (global or absolute) maximum (a \(y\)-value that it never goes above) or a (global or absolute) minimum (a \(y\)-value that it never goes below). That max or min is the vertex of the parabola. On the flip side, in whichever direction the parabola opens, it will go on forever and ever that way, which brings us to the subject of end behavior. End behavior means the behavior of a function as \(x\) goes off to the right forever to infinity, and as \(x\) goes off to the left forever to negative infinity. We can use symbols and words to describe end behavior. For example, the first graph above goes up forever as we send \(x\) off to positive infinity, and also up forever as we send \(x\) to negative infinity. We write:

• as \( x \rightarrow \infty\), \( f(x) \rightarrow \infty\)
• as \(x \rightarrow -\infty\), \( f(x) \rightarrow \infty\)

Similarly, for the downward-opening second graph,

• as \(x \rightarrow \infty\), \( f(x) \rightarrow -\infty\)
• as \(x \rightarrow -\infty\), \( f(x) \rightarrow -\infty\)

This means that the range of a quadratic function will have one infinity symbol in it. For example, the first function hits all \(y\)-values from \(2\) onward, so the range is \( [2, \infty) \). The second function has range \( (-\infty, 3] \). What is the range of the third function?

While we're here, notice also that there are three possibilities when it comes to \(x\)-intercepts. The first example never touches the \(x\)-axis, so it has no real roots. The second example crosses the \(x\)-axis in two distinct locations, so it has distinct real roots. The third example bounces off the \(x\)-axis without crossing over, so it has what is called a repeated root. We'll see why it's called repeated shortly, when we learn how to find these roots algebraically.

Example \(\PageIndex{1}\)

Tell whether the functions are quadratic functions, and if necessary, FOIL them into standard form.

1. \( f(x) = 11x^2 + 12x + 13 \)
2. \( g(x) = \dfrac{x^2 + x + 1}{x-1} \)
3. \( h(x) = x^2 - x^3 \)
4. \( p(x) = \frac{1}{4} + \frac{1}{2}x + x^2\)
5. \( q(x) = (x+3)^2 \)
6. \( A(r) = \pi r^2 \)
7. \( C(x) = 3000 +30x - 0.5x^2 \)

Solution

1. Yes.
2. No.
3. No, this is degree-three.
4. Yes, you could order the terms from largest power of \(x\) to smallest, if you want.
5. Yes, \(q(x) = x^2 + 6x + 9 \).
6. Yes, the variable \(r\) is squared. You could view this in standard form as \(A(r) = \pi r^2 + 0 r + 0 \) meaning \(a = \pi, b = 0, c = 0\).
7. Yes.

Exercise \(\PageIndex{1}\)

Tell whether the functions are quadratic functions, and if necessary, rewrite them in standard form.

1. \( f(x) = \sqrt{ x^2 + x + 1} \)
2. \( g(x) = 3x^2 - 1 \)
3. \( h(t) = 100 - 4.9 t^2 \)
4. \( p(x) = 2x^2 + 4(x-1)^2 \)
5. \( q(x) = (x + 1)(x - 2) \)
6. \( r(x) = \dfrac{x^2 + 6x + 1}{x - 2} \)
7. \( C(x) = 200 + 40x - x^{\frac{3}{2}} \)

Answer

1. No.
2. Yes.
3. Yes.
4. Yes, \( p(x) = 6x^2 - 8x + 4 \).
5. Yes, \(q(x) = x^2 - x - 2 \).
6. No.
7. No.

Vertex Form

The vertex form of a quadratic function is very useful, because knowing the location of the vertex helps you sketch a graph, tells you where the function's max or min is, and in combination with knowledge of which way the parabola opens, tells you everything you need to assess range and narrow down root possibilities. A quadratic function is written in vertex form if it looks like this:

\[ f(x) = a(x-h)^2 + k \notag \]

where \(a, h,\) and \(k\) are constants. We know here that if \(a > 0\), the parabola opens up, and if \( a<0\), it opens down. We also know that \( (h,k)\) is the location of the vertex.

(Why?? Remember transformations? To go from a "basic" function \(g(x) = x^2 \) to the function \(f(x) = a(x-h)^2 + k\), we see that several transformations happened: a vertical stretch or shrink, then a shift to the right by \(h\), and a shift up by \(k\). If the vertex started at the origin, \( (0,0)\) and was shifted right by \(h\) and up by \(k\), it's now living at \( (h,k) \)!)

To identify the vertex, simply make sure the function matches the vertex form above, and then read off the values of \(h\) and \(k\).

Exercise \(\PageIndex{2}\)

Identify the vertex.

1. \( f(x) = 3(x-1)^2 + 2 \)
2. \( f(x) = (x+2)^2 - 16 \)
3. \( f(x) = x^2 + 1 \)
4. \( f(x) = \frac{1}{2}(x + 10)^2 \)

Answer

1. Here \(h = 1\) and \( k = 2\), so the vertex is at \( (1,2)\).
2. We could rewrite this slightly as \( f(x) = (x - (-2))^2 + (-16) \) so that it's easy to identify \( (-2, -16) \).
3. We could rewrite this as \( f(x) = (x-0)^2 + 1 \) to identify \( (0, 1) \).
4. Similarly, we notice \( f(x) = \frac{1}{2}(x-(-10))^2 + 0 \) to identify \( (-10, 0)\).

That's great and all, but what if the vertex form isn't obvious or already given? Then we have to find it ourselves. There's a process for this.

To convert from standard form \(f(x) = ax^2 + bx + c \) into vertex form \( f(x) = a(x-h)^2 + k \),

1. If \(a \neq 1\), factor it out like so: \( f(x) = a \left( x^2 + \frac{b}{a} x + \frac{c}{a} \right) \).
2. Working inside the parentheses, complete the square on the first two terms, \(x^2 + \frac{b}{a} x \), making sure to both add and subtract the necessary term inside the parentheses.
3. If necessary, distribute the factored-out \(a\) back onto the constant term to finally reach the form \( f(x) = a(x-h)^2 + k \) for some \(h\) and \(k\).

Once you know the location of the vertex and whether the parabola opens up or down, you can easily determine the range. Let's include that information in our examples.

Example \(\PageIndex{2}\)

1. Convert \( f(x) = x^2 -2x + 3 \) into vertex form. Then identify the vertex and the range of the function.

2. Convert \( f(x) = -3x^2 -6x + 3 \) into vertex form. Then identify the vertex and the range of the function.

Solution

1. Here, \(a = 1\) so we don't have to worry about that. We can go straight to completing the square. We take \(b = -2\), divide by 2, and square the result to get \(1\). Then we sneakily add zero to the function by adding and subtracting \(1\).

\[ f(x) = x^2 - 2x \quad \quad + 3 \quad \longrightarrow \quad f(x) = x^2 - 2x \textcolor{red}{+ 1} + 3 \textcolor{red}{-1} \notag \]

Now the \(x^2 - 2x + 1\) has been engineered to factor as \( (x-1)^2\), so we finish simplifying:

\[ f(x) = (x-1)^2 + 2 \notag \]

We can see the vertex is \( (1, 2)\), and since the leading coefficient is positive, the parabola must open upward. That means \(y = 2\) is the lowest my function ever goes. The range must be \( [2, \infty) \).

2. In this case, \(a \neq 1\), so we will have to carefully push him around throughout. First, we get him out of the way for now by writing \( f(x) = -3( x^2 + 2x - 1)\), and we focus on the inside of the parentheses. If I want to complete the square on \( x^2 + 2x\), I just take the \(2\), divide by 2, and square the result to get \(1\). Still inside the parentheses, I add and subtract that result.

\[ f(x) = -3 (x^2 + 2x \textcolor{red}{+1 } - 1 \textcolor{red}{-1}) \quad \longrightarrow \quad f(x) = -3( (x+1)^2 - 2 )\notag \]

Now, I just distribute that \(3\) we had factored out onto the perfect square term and onto the constant \(-2\) so that I fully match the vertex form:

\[ f(x) = -3(x+1)^2 + 6. \notag \]

The vertex is \( (-1, 6) \). Since the leading coefficient \(-3\) was negative, the parabola opens downward. That means the \(y = 6\) height is the maximum function value, so my range is \( (-\infty, 6] \).

Exercise \(\PageIndex{3}\)

Convert to vertex form. Then identify the vertex and range of the function.

1. \( f(x) = x^2 - 4x + 8 \)
2. \( f(x) = 2x^2 + 12x + 17 \)
3. \( f(x) = -2x^2 -20x -50 \)

Answer

1. \( f(x) = (x-2)^2 + 4 \). The vertex is \( (2, 4) \) and the range is \( [4,\infty) \).
2. \( f(x) = 2(x+3)^2 - 1 \). The vertex is \( (-3,-1) \) and the range is \( [-1, \infty) \).
3. \( f(x) = -2(x+5)^2 + 0 \). The vertex is \( (-5,0) \) and the range is \( (-\infty, 0] \).

Finding Roots of Quadratic Functions

The roots of a quadratic function are just the \(x\)-values of the \(x\)-intercepts. To be on the \(x\)-axis, I must have a \(y\)-value of 0, so I am looking for the locations at which
\[ f(x)= ax^2 + bx + c = 0. \notag \]
Whoop, that's just giving me a quadratic equation, \(ax^2 + bx + c = 0\). We've solved the heck out of those already, using various techniques.

Now, when you go about this process, there are three scenarios that could arise. Let me demonstrate them with an example:

Example \(\PageIndex{3}\)

Find all real roots of the quadratic functions:

1. \( f(x) = x^2 + 5x + 6 \)
2. \( f(x) = 2x^2 + 4x + 2 \)
3. \( f(x) = x^2 -4x + 10 \)

Solution

1. Always just set the function equal to 0 as the first step. I solve the resulting quadratic equation for \(x\):

\[ x^2 + 5x + 6 = 0 \quad \longrightarrow \quad (x+2)(x+3) = 0 \quad \longrightarrow \quad x = -2, -3\notag \]

Here, my quadratic factored nicely into two distinct factors, resulting in two different real number roots. This guy looks like this:

2. Notice that each term has a \(2\) in it, so I try a factoring method again:

\[ 2x^2 + 4x + 2 = 0 \quad \longrightarrow \quad 2(x^2 + 2x + 1) = 0 \quad \longrightarrow \quad 2(x+1)^2 = 0\notag \]

Notice that this one turned out to have two copies of the same factor, \( (x+1)\). Aka, the factor was repeated. In this case, solving for \(x\) yields only one value, \( x = -1\), but we call this a repeated root, or a root with multiplicity 2, because the factor that it came from appeared twice. Sometimes I write this answer as "\( x = -1, -1 \)" so I remember what's going on. This graph looks like this, bouncing off the \(x\)-axis:

3. I set this guy equal to 0 as usual, \( x^2 - 4x + 10 = 0\), but I don't see an easy way to factor him at the moment. I don't waste too much time thinking about it before giving the quadratic formula a shot. I indentify \(a = 1, b = -4, c = 10\) and plug in...

\[ x = \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{ 4 \pm \sqrt{ 16 - 4(1)(10)}}{2} = \dfrac{ 4 \pm \sqrt{-24}}{2} \notag \]

Ope, companyyyyyy halt! The discriminant, \(b^2 - 4ac \), turned out to be negative here. There are no real roots! If we pushed on into the complex numbers, the solution to this equation would be a complex conjugate pair. Look what happens on the graph... the parabola doesn't hit the \(x\)-axis at all!

Now you practice finding roots! You can use any method you like for this, including the \(ac\) method, completing the square, grouping, whatever.

Exercise \(\PageIndex{4}\)

Find any and all real roots of the quadratic functions.

1. \( f(x) = x^2 - 2x - 8 \)
2. \( f(x) = x^2 - 6x + 9 \)
3. \( f(x) = 3x^2 - x - 2 \)
4. \( f(x) = 4x^2 -12x + 8 \)
5. \( f(x) = x^2 - x + 2 \)

Answer

1. \( x = -2, 4 \)
2. \( x = 3, 3 \)
3. \( x = -\frac{2}{3}, 1 \)
4. \( x = 1, 2 \)
5. No real roots.

Sketching Graphs of Quadratic Functions

To sketch the graph of a quadratic function,

1. Find the vertex form and plot the vertex. Note whether the parabola should open up or down.
2. Find a few more points by plugging in different \(x\)-values and computing the function values. You can also choose to find the roots instead of random points.
3. Using knowledge of symmetry and how parabolas look, fit a parabola to the the plotted points smoothly.

Or, use knowledge of transformations to sketch the graph, starting with the most basic \( f(x) = x^2\) and performing the appropriate transformations.

Example \(\PageIndex{4}\)

Sketch the graphs of the quadratic functions.

1. \( f(x) = x^2 -2x + 3 \)
2. \( f(x) = -3x^2 -6x + 3 \)
3. \( f(x) = x^2 - 2x - 8 \)
4. \( f(x) = 3x^2 - x - 2 \)

Solution

1.	2.	3.	4.
We've already found the vertex form in a previous example: \( f(x) = (x-1)^2 + 2\). Plot the vertex \( (1,2) \). I compute a point on either side of the vertex: \[ f(0) = 3 \: \longrightarrow \: (0,3) \notag \] \[ f(3) = 6 \: \longrightarrow \: (3,6) \notag \] Connect the dots knowing that the parabola opens upward forever.	We have the vertex form: \( f(x) = -3(x+1)^2 + 6\). Plot the vertex \( (-1, 6) \). Note that this parabola needs to open downward. I compute an easy point: \[ f(0) = 3 \: \longrightarrow \: (0,3) \notag \] and use symmetry to deduce another point, \( (-2, 3) \). Connect the dots.	I see the roots of this function are \( x = -2, 4 \), which gives me the two \(x\)-intercept points, \( (-2,0)\) and \( (4,0)\). I still need the vertex to draw a decent graph. Completing the square gets me to the vertex form \[ f(x) = (x-1)^2 - 9\notag \] so I see the vertex is \( (1,-9)\) and the parabola should open upwards. Connect the dots.	We found the roots of this function to be \( x = -\frac{2}{3}, 1\) so I have two points, \( \left( -\frac{2}{3}, 0 \right)\) and \( (1,0) \). I complete the square to get the vertex form \[ f(x) = 3\left( x - \frac{1}{6}\right)^2 - \frac{25}{12}\notag \] so I see the vertex is \( \left( \frac{1}{6}, - \frac{25}{12} \right) \). Noting that \( \frac{1}{6}\) is less than 0.2, and \(\frac{25}{12}\) is just a hair more than 2, I sketch the point and connect the dots.

Often, though, all you really need is to know exactly what \( f(x) = x^2 \) looks like, and then be able to recognize transformations.

Example \(\PageIndex{5}\)

Without performing any real calculations, use knowledge of transformations to sketch the graphs of the quadratic functions, starting with the graph of \(f(x) = x^2 \):

1. \( g(x) = x^2 - 2 \)
2. \( g(x) = (x - 1)^2 + 3 \)
3. \( g(x) = -x^2 + 4 \)

Solution

1. We recognize that the transformation "\( x^2 \: \longrightarrow \: x^2 - 2\)" is a shift down by 2 units. We take some signpost points on the original graph, like \( (0,0)\), \( (1,1) \), \( (2,4)\), and imagine sliding them down by 2 units to help us sketch.

2. We recognize the transformations

\[ x^2 \quad \longrightarrow \quad (x-1)^2 \quad \longrightarrow \quad (x-1)^2 + 3 \notag \]

are a shift right by 1 and shift up by 3. We take a few signpost points and perform those shifts on them to help us sketch. For example, \( (0,0) \) becomes \( (1,3) \).

3. We recognize that the transformations here were first a reflection over the \(x\)-axis, and then a shift up by 4. We can imagine an intermediate step if that helps us sketch the graph, shown in dotted red below.

Exercise \(\PageIndex{5}\)

Using either of the above methods, sketch the graphs of the quadratic functions.

1. \( f(x) = -(x-3)^2 + 2 \)
2. \( f(x) = x^2 + 2x - 4 \)
3. \( f(x) = x^2 + 1 \)

Answer

1.	2.	3.

Now, be sure to do the exercises section, because we're going to see some real-life application problems over there!