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# Quadratic Functions and their Graphs

### Learning Objectives

• Roots or zeros of a quadratic function
• Characteristics of a parabola
• vertex
• axis of symmetry
• x/y-intercepts
• Classifying solutions to quadratic equations
• The discriminant
Curved antennas, such as the ones shown in the photo, are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function.
An array of satellite dishes. (credit: Matthew Colvin de Valle, Flickr)

## Characteristics of Parabolas

The graph of a quadratic function is a U-shaped curve called a parabola. One important feature of the graph is that it has an extreme point, called the vertex. If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value. In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry. The y-intercept is the point at which the parabola crosses the y-axis. The x-intercepts are the points at which the parabola crosses the x-axis. If they exist, the x-intercepts represent the zeros, or roots, of the quadratic function, the values of x at which = 0.

### Example: Identifying the Characteristics of a Parabola

Determine the vertex, axis of symmetry, zeros, and y-intercept of the parabola shown below.

Answer: The vertex is the turning point of the graph. We can see that the vertex is at (3, 1). Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is = 3. This parabola does not cross the x-axis, so it has no zeros. It crosses the y-axis at (0, 7) so this is the y-intercept.

### Try It

[ohm_question]147099[/ohm_question]

### General and Standard Forms of Quadratic Functions

The general form of a quadratic function presents the function in the form

$f\left(x\right)=a{x}^{2}+bx+c$

where ab, and c are real numbers and $a\ne 0$. If $a>0$, the parabola opens upward. If $a<0$, the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry. The axis of symmetry is defined by $x=-\frac{b}{2a}$. If we use the quadratic formula, $x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$, to solve $a{x}^{2}+bx+c=0$ for the x-intercepts, or zeros, we find the value of x halfway between them is always $x=-\frac{b}{2a}$, the equation for the axis of symmetry. The figure below shows the graph of the quadratic function written in general form as $y={x}^{2}+4x+3$. In this form, $a=1,\text{ }b=4$, and $c=3$. Because $a>0$, the parabola opens upward. The axis of symmetry is $x=-\frac{4}{2\left(1\right)}=-2$. This also makes sense because we can see from the graph that the vertical line $x=-2$ divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, $\left(-2,-1\right)$. The x-intercepts, those points where the parabola crosses the x-axis, occur at $\left(-3,0\right)$ and $\left(-1,0\right)$. The standard form of a quadratic function presents the function in the form

$f\left(x\right)=a{\left(x-h\right)}^{2}+k$

where $\left(h,\text{ }k\right)$ is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.

### Given a quadratic function in general form, find the vertex of the parabola.

One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, (k), and where it occurs, (x). If we are given the general form of a quadratic function:

$f(x)=ax^2+bx+c$

We can define the vertex, $(h,k)$, by doing the following:
• Identify ab, and c.
• Find h, the x-coordinate of the vertex, by substituting a and b into $h=-\frac{b}{2a}$.
• Find k, the y-coordinate of the vertex, by evaluating $k=f\left(h\right)=f\left(-\frac{b}{2a}\right)$

### Example: Finding the Vertex of a Quadratic Function

Find the vertex of the quadratic function $f\left(x\right)=2{x}^{2}-6x+7$. Rewrite the quadratic in standard form (vertex form).

Answer: The horizontal coordinate of the vertex will be at

$\begin{array}{c}h=-\frac{b}{2a}\hfill \\ \text{ }=-\frac{-6}{2\left(2\right)}\hfill \\ \text{ }=\frac{6}{4}\hfill \\ \text{ }=\frac{3}{2}\hfill \end{array}$

The vertical coordinate of the vertex will be at

$\begin{array}{c}k=f\left(h\right)\hfill \\ \text{ }=f\left(\frac{3}{2}\right)\hfill \\ \text{ }=2{\left(\frac{3}{2}\right)}^{2}-6\left(\frac{3}{2}\right)+7\hfill \\ \text{ }=\frac{5}{2}\hfill \end{array}$

Rewriting into standard form, the stretch factor will be the same as the $a$ in the original quadratic.

$\begin{array}{c}f\left(x\right)=a{x}^{2}+bx+c\hfill \\ f\left(x\right)=2{x}^{2}-6x+7\hfill \end{array}$

Using the vertex to determine the shifts,

$f\left(x\right)=2{\left(x-\frac{3}{2}\right)}^{2}+\frac{5}{2}$

### Try It

Given the equation $g\left(x\right)=13+{x}^{2}-6x$, write the equation in general form and then in standard form.

Answer: $g\left(x\right)={x}^{2}-6x+13$ in general form; $g\left(x\right)={\left(x - 3\right)}^{2}+4$ in standard form

In this section, we will investigate quadratic functions further, including solving problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree polynomial functions, so they provide a good opportunity for a detailed study of function behavior.

## Classifying Solutions to Quadratic Equations

Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the y-intercept of a quadratic by evaluating the function at an input of zero, and we find the x-intercepts at locations where the output is zero. Notice that the number of x-intercepts can vary depending upon the location of the graph.
Number of x-intercepts of a parabola
Mathematicians also define x-intercepts as roots of the quadratic function.

### How To: Given a quadratic function $f\left(x\right)$, find the y- and x-intercepts.

1. Evaluate $f\left(0\right)$ to find the y-intercept.
2. Solve the quadratic equation $f\left(x\right)=0$ to find the x-intercepts.

### Example: Finding the y- and x-Intercepts of a Parabola

Find the y- and x-intercepts of the quadratic $f\left(x\right)=3{x}^{2}+5x - 2$.

Answer: We find the y-intercept by evaluating $f\left(0\right)$.

$\begin{array}{c}f\left(0\right)=3{\left(0\right)}^{2}+5\left(0\right)-2\hfill \\ \text{ }=-2\hfill \end{array}$

So the y-intercept is at $\left(0,-2\right)$. For the x-intercepts, or roots, we find all solutions of $f\left(x\right)=0$.

$0=3{x}^{2}+5x - 2$

In this case, the quadratic can be factored easily, providing the simplest method for solution.

$0=\left(3x - 1\right)\left(x+2\right)$ $\begin{array}{c}0=3x - 1\hfill & \hfill & \hfill & \hfill & 0=x+2\hfill \\ x=\frac{1}{3}\hfill & \hfill & \text{or}\hfill & \hfill & x=-2\hfill \end{array}$

So the roots are at $\left(\frac{1}{3},0\right)$ and $\left(-2,0\right)$.

#### Analysis of the Solution

By graphing the function, we can confirm that the graph crosses the y-axis at $\left(0,-2\right)$. We can also confirm that the graph crosses the x-axis at $\left(\frac{1}{3},0\right)$ and $\left(-2,0\right)$.

In Example: Finding the y- and x-Intercepts of a Parabola, the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.

### How To: Given a quadratic function, find the x-intercepts by rewriting in standard form.

1. Substitute a and b into $h=-\frac{b}{2a}$.
2. Substitute xh into the general form of the quadratic function to find k.
3. Rewrite the quadratic in standard form using h and k.
4. Solve for when the output of the function will be zero to find the x-intercepts.

### Example: Finding the Roots of a Parabola

Find the x-intercepts of the quadratic function $f\left(x\right)=2{x}^{2}+4x - 4$.

Answer: We begin by solving for when the output will be zero.

$0=2{x}^{2}+4x - 4$

Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.

$f\left(x\right)=a{\left(x-h\right)}^{2}+k$

We know that = 2. Then we solve for h and k.

$\begin{array}{c}h=-\frac{b}{2a}\hfill & \hfill & \hfill & k=f\left(-1\right)\hfill \\ \text{ }=-\frac{4}{2\left(2\right)}\hfill & \hfill & \hfill & \text{ }=2{\left(-1\right)}^{2}+4\left(-1\right)-4\hfill \\ \text{ }=-1\hfill & \hfill & \hfill & \text{ }=-6\hfill \end{array}$

So now we can rewrite in standard form.

$f\left(x\right)=2{\left(x+1\right)}^{2}-6$

We can now solve for when the output will be zero.

$\begin{array}{c}0=2{\left(x+1\right)}^{2}-6\hfill \\ 6=2{\left(x+1\right)}^{2}\hfill \\ 3={\left(x+1\right)}^{2}\hfill \\ x+1=\pm \sqrt{3}\hfill \\ x=-1\pm \sqrt{3}\hfill \end{array}$

The graph has x-intercepts at $\left(-1-\sqrt{3},0\right)$ and $\left(-1+\sqrt{3},0\right)$.

#### Analysis of the Solution

We can check our work by graphing the given function on a graphing utility and observing the roots.

### Try It

Find the y-intercept for the function $g\left(x\right)=13+{x}^{2}-6x$.

## Complex Roots

Now you will hopefully begin to understand why we introduced complex numbers at the beginning of this module. Consider the following function: $f(x)=x^2+2x+3$, and it's graph below: Does this function have roots? It's probably obvious that this function does not cross the x-axis, therefore it doesn't have any x-intercepts. Recall that the x-intercepts of a function are found by setting the function equal to zero:

$x^2+2x+3=0$

In the next example, we will solve this equation.  You will see that there are roots, but they are not x-intercepts because the function does not contain (x,y) pairs that are on the x-axis.  We call these complex roots. By setting the function equal to zero and using the quadratic formula to solve, you will see that the roots contain complex numbers:

$x^2+2x+3=0$

### Example

Find the x-intercepts of the quadratic function. $f(x)=x^2+2x+3$

Answer: The x-intercepts of the function $f(x)=x^2+2x+3$ are found by setting it equal to zero, and solving for x since the y values of the x-intercepts are zero. First, identify a, b, c.

$\begin{array}{ccc}x^2+2x+3=0\\a=1,b=2,c=3\end{array}$

Substitute these values into the quadratic formula.

$\begin{array}{c}x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}\\=\frac{-2\pm \sqrt{{2}^{2}-4(1)(3)}}{2(1)}\\=\frac{-2\pm \sqrt{4-12}}{2} \\=\frac{-2\pm \sqrt{-8}}{2}\\\frac{-2\pm 2i\sqrt{2}}{2} \\-1\pm i\sqrt{2}=-1+\sqrt{2},-1-\sqrt{2}\end{array}$

The solutions to this equations are complex, therefore there are no x-intercepts for the function $f(x)=x^2+2x+3$ in the set of real numbers that can be plotted on the Cartesian Coordinate plane. The graph of the function is plotted on the Cartesian Coordinate plane below:
Graph of quadratic function with no x-intercepts in the real numbers.
Note how the graph does not cross the x-axis, therefore there are no real x-intercepts for this function.

### The Discriminant

The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions. When we consider the discriminant, or the expression under the radical, ${b}^{2}-4ac$, it tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. In turn, we can then determine whether a quadratic function has real or complex roots. The table below relates the value of the discriminant to the solutions of a quadratic equation.
Value of Discriminant Results
${b}^{2}-4ac=0$ One repeated rational solution
${b}^{2}-4ac>0$, perfect square Two rational solutions
${b}^{2}-4ac>0$, not a perfect square Two irrational solutions
${b}^{2}-4ac<0$ Two complex solutions

### A General Note: The Discriminant

For $a{x}^{2}+bx+c=0$, where $a$, $b$, and $c$ are real numbers, the discriminant is the expression under the radical in the quadratic formula: ${b}^{2}-4ac$. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect.

### Example

Use the discriminant to find the nature of the solutions to the following quadratic equations:
1. ${x}^{2}+4x+4=0$
2. $8{x}^{2}+14x+3=0$
3. $3{x}^{2}-5x - 2=0$
4. $3{x}^{2}-10x+15=0$

Answer: Calculate the discriminant ${b}^{2}-4ac$ for each equation and state the expected type of solutions.

1. ${x}^{2}+4x+4=0$${b}^{2}-4ac={\left(4\right)}^{2}-4\left(1\right)\left(4\right)=0$. There will be one repeated rational solution.
2. $8{x}^{2}+14x+3=0$${b}^{2}-4ac={\left(14\right)}^{2}-4\left(8\right)\left(3\right)=100$. As $100$ is a perfect square, there will be two rational solutions.
3. $3{x}^{2}-5x - 2=0$${b}^{2}-4ac={\left(-5\right)}^{2}-4\left(3\right)\left(-2\right)=49$. As $49$ is a perfect square, there will be two rational solutions.
4. $3{x}^{2}-10x+15=0$${b}^{2}-4ac={\left(-10\right)}^{2}-4\left(3\right)\left(15\right)=-80$. There will be two complex solutions.

We have seen that a quadratic equation may have two real solutions, one real solution, or two complex solutions. Let’s summarize how the discriminant affects the evaluation of $\sqrt{{{b}^{2}}-4ac}$, and how it helps to determine the solution set.
• If $b^{2}-4ac>0$, then the number underneath the radical will be a positive value. You can always find the square root of a positive, so evaluating the quadratic formula will result in two real solutions (one by adding the positive square root, and one by subtracting it).
• If $b^{2}-4ac=0$, then you will be taking the square root of 0, which is 0. Since adding and subtracting 0 both give the same result, the "$\pm[/late]" portion of the formula doesn't matter. There will be one real repeated solution. • If [latex]b^{2}-4ac<0$, then the number underneath the radical will be a negative value. Since you cannot find the square root of a negative number using real numbers, there are no real solutions. However, you can use imaginary numbers. You will then have two complex solutions, one by adding the imaginary square root and one by subtracting it.

### Example

Use the discriminant to determine how many and what kind of solutions the quadratic equation $x^{2}-4x+10=0$ has.

Answer: Evaluate $b^{2}-4ac$. First note that $a=1,b=−4$, and $c=10$.

$\begin{array}{c}b^{2}-4ac\\\left(-4\right)^{2}-4\left(1\right)\left(10\right)\end{array}$

The result is a negative number. The discriminant is negative, so the quadratic equation has two complex solutions.

$16–40=−24$

The quadratic equation $x^{2}-4x+10=0$ has two complex solutions.

## Important Terms

axis of symmetry
a vertical line drawn through the vertex of a parabola around which the parabola is symmetric; it is defined by $x=-\frac{b}{2a}$.
general form of a quadratic function
the function that describes a parabola, written in the form $f\left(x\right)=a{x}^{2}+bx+c$, where ab, and c are real numbers and $a\ne 0$.
standard form of a quadratic function
the function that describes a parabola, written in the form $f\left(x\right)=a{\left(x-h\right)}^{2}+k$, where $\left(h,\text{ }k\right)$ is the vertex.
discriminant
the value under the radical in the quadratic formula, $b^2-4ac$, which tells whether the quadratic has real or complex roots
vertex
the point at which a parabola changes direction, corresponding to the minimum or maximum value of the quadratic function
zeros
in a given function, the values of x at which y = 0, also called roots