Quadratic Functions and their Graphs
Learning Objectives
- Quadratic Functions
- 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
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 y = 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 x = 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[latex]f\left(x\right)=a{x}^{2}+bx+c[/latex]
where a, b, and c are real numbers and [latex]a\ne 0[/latex]. If [latex]a>0[/latex], the parabola opens upward. If [latex]a<0[/latex], 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 [latex]x=-\frac{b}{2a}[/latex]. If we use the quadratic formula, [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex], to solve [latex]a{x}^{2}+bx+c=0[/latex] for the x-intercepts, or zeros, we find the value of x halfway between them is always [latex]x=-\frac{b}{2a}[/latex], the equation for the axis of symmetry. The figure below shows the graph of the quadratic function written in general form as [latex]y={x}^{2}+4x+3[/latex]. In this form, [latex]a=1,\text{ }b=4[/latex], and [latex]c=3[/latex]. Because [latex]a>0[/latex], the parabola opens upward. The axis of symmetry is [latex]x=-\frac{4}{2\left(1\right)}=-2[/latex]. This also makes sense because we can see from the graph that the vertical line [latex]x=-2[/latex] 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, [latex]\left(-2,-1\right)[/latex]. The x-intercepts, those points where the parabola crosses the x-axis, occur at [latex]\left(-3,0\right)[/latex] and [latex]\left(-1,0\right)[/latex].
The standard form of a quadratic function presents the function in the form
[latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex]
where [latex]\left(h,\text{ }k\right)[/latex] 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:[latex]f(x)=ax^2+bx+c[/latex]
We can define the vertex, [latex](h,k)[/latex], by doing the following:- Identify a, b, and c.
- Find h, the x-coordinate of the vertex, by substituting a and b into [latex]h=-\frac{b}{2a}[/latex].
- Find k, the y-coordinate of the vertex, by evaluating [latex]k=f\left(h\right)=f\left(-\frac{b}{2a}\right)[/latex]
Example: Finding the Vertex of a Quadratic Function
Find the vertex of the quadratic function [latex]f\left(x\right)=2{x}^{2}-6x+7[/latex]. Rewrite the quadratic in standard form (vertex form).Answer: The horizontal coordinate of the vertex will be at
[latex]\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}[/latex]
The vertical coordinate of the vertex will be at[latex]\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}[/latex]
Rewriting into standard form, the stretch factor will be the same as the [latex]a[/latex] in the original quadratic.[latex]\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}[/latex]
Using the vertex to determine the shifts,[latex]f\left(x\right)=2{\left(x-\frac{3}{2}\right)}^{2}+\frac{5}{2}[/latex]
Try It
Given the equation [latex]g\left(x\right)=13+{x}^{2}-6x[/latex], write the equation in general form and then in standard form.Answer: [latex]g\left(x\right)={x}^{2}-6x+13[/latex] in general form; [latex]g\left(x\right)={\left(x - 3\right)}^{2}+4[/latex] in standard form
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 parabolaHow To: Given a quadratic function [latex]f\left(x\right)[/latex], find the y- and x-intercepts.
- Evaluate [latex]f\left(0\right)[/latex] to find the y-intercept.
- Solve the quadratic equation [latex]f\left(x\right)=0[/latex] to find the x-intercepts.
Example: Finding the y- and x-Intercepts of a Parabola
Find the y- and x-intercepts of the quadratic [latex]f\left(x\right)=3{x}^{2}+5x - 2[/latex].Answer: We find the y-intercept by evaluating [latex]f\left(0\right)[/latex].
[latex]\begin{array}{c}f\left(0\right)=3{\left(0\right)}^{2}+5\left(0\right)-2\hfill \\ \text{ }=-2\hfill \end{array}[/latex]
So the y-intercept is at [latex]\left(0,-2\right)[/latex]. For the x-intercepts, or roots, we find all solutions of [latex]f\left(x\right)=0[/latex].[latex]0=3{x}^{2}+5x - 2[/latex]
In this case, the quadratic can be factored easily, providing the simplest method for solution.[latex]0=\left(3x - 1\right)\left(x+2\right)[/latex] [latex]\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}[/latex]
So the roots are at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].Analysis of the Solution
By graphing the function, we can confirm that the graph crosses the y-axis at [latex]\left(0,-2\right)[/latex]. We can also confirm that the graph crosses the x-axis at [latex]\left(\frac{1}{3},0\right)[/latex] and [latex]\left(-2,0\right)[/latex].
How To: Given a quadratic function, find the x-intercepts by rewriting in standard form.
- Substitute a and b into [latex]h=-\frac{b}{2a}[/latex].
- Substitute x = h into the general form of the quadratic function to find k.
- Rewrite the quadratic in standard form using h and k.
- 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 [latex]f\left(x\right)=2{x}^{2}+4x - 4[/latex].Answer: We begin by solving for when the output will be zero.
[latex]0=2{x}^{2}+4x - 4[/latex]
Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.[latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex]
We know that a = 2. Then we solve for h and k.[latex]\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}[/latex]
So now we can rewrite in standard form.[latex]f\left(x\right)=2{\left(x+1\right)}^{2}-6[/latex]
We can now solve for when the output will be zero.[latex]\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}[/latex]
The graph has x-intercepts at [latex]\left(-1-\sqrt{3},0\right)[/latex] and [latex]\left(-1+\sqrt{3},0\right)[/latex].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 [latex]g\left(x\right)=13+{x}^{2}-6x[/latex].Answer: y-intercept at (0, 13)
Complex Roots
Now you will hopefully begin to understand why we introduced complex numbers at the beginning of this module. Consider the following function: [latex]f(x)=x^2+2x+3[/latex], 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:
[latex]x^2+2x+3=0[/latex]
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:[latex]x^2+2x+3=0[/latex]
Example
Find the x-intercepts of the quadratic function. [latex]f(x)=x^2+2x+3[/latex]Answer: The x-intercepts of the function [latex]f(x)=x^2+2x+3[/latex] 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.
[latex]\begin{array}{ccc}x^2+2x+3=0\\a=1,b=2,c=3\end{array}[/latex]
Substitute these values into the quadratic formula.[latex]\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}[/latex]
The solutions to this equations are complex, therefore there are no x-intercepts for the function [latex]f(x)=x^2+2x+3[/latex] 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.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, [latex]{b}^{2}-4ac[/latex], 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 |
|---|---|
| [latex]{b}^{2}-4ac=0[/latex] | One repeated rational solution |
| [latex]{b}^{2}-4ac>0[/latex], perfect square | Two rational solutions |
| [latex]{b}^{2}-4ac>0[/latex], not a perfect square | Two irrational solutions |
| [latex]{b}^{2}-4ac<0[/latex] | Two complex solutions |
A General Note: The Discriminant
For [latex]a{x}^{2}+bx+c=0[/latex], where [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are real numbers, the discriminant is the expression under the radical in the quadratic formula: [latex]{b}^{2}-4ac[/latex]. 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:- [latex]{x}^{2}+4x+4=0[/latex]
- [latex]8{x}^{2}+14x+3=0[/latex]
- [latex]3{x}^{2}-5x - 2=0[/latex]
- [latex]3{x}^{2}-10x+15=0[/latex]
Answer: Calculate the discriminant [latex]{b}^{2}-4ac[/latex] for each equation and state the expected type of solutions.
- [latex]{x}^{2}+4x+4=0[/latex][latex]{b}^{2}-4ac={\left(4\right)}^{2}-4\left(1\right)\left(4\right)=0[/latex]. There will be one repeated rational solution.
- [latex]8{x}^{2}+14x+3=0[/latex][latex]{b}^{2}-4ac={\left(14\right)}^{2}-4\left(8\right)\left(3\right)=100[/latex]. As [latex]100[/latex] is a perfect square, there will be two rational solutions.
- [latex]3{x}^{2}-5x - 2=0[/latex][latex]{b}^{2}-4ac={\left(-5\right)}^{2}-4\left(3\right)\left(-2\right)=49[/latex]. As [latex]49[/latex] is a perfect square, there will be two rational solutions.
- [latex]3{x}^{2}-10x+15=0[/latex][latex]{b}^{2}-4ac={\left(-10\right)}^{2}-4\left(3\right)\left(15\right)=-80[/latex]. There will be two complex solutions.
- If [latex]b^{2}-4ac>0[/latex], 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 [latex]b^{2}-4ac=0[/latex], then you will be taking the square root of 0, which is 0. Since adding and subtracting 0 both give the same result, the "[latex]\pm[/late]" portion of the formula doesn't matter. There will be one real repeated solution.
- If [latex]b^{2}-4ac<0[/latex], 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 [latex]x^{2}-4x+10=0[/latex] has.Answer: Evaluate [latex]b^{2}-4ac[/latex]. First note that [latex]a=1,b=−4[/latex], and [latex]c=10[/latex].
[latex]\begin{array}{c}b^{2}-4ac\\\left(-4\right)^{2}-4\left(1\right)\left(10\right)\end{array}[/latex]
The result is a negative number. The discriminant is negative, so the quadratic equation has two complex solutions.[latex]16–40=−24[/latex]
Answer
The quadratic equation [latex]x^{2}-4x+10=0[/latex] 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 [latex]x=-\frac{b}{2a}[/latex].
- general form of a quadratic function
- the function that describes a parabola, written in the form [latex]f\left(x\right)=a{x}^{2}+bx+c[/latex], where a, b, and c are real numbers and [latex]a\ne 0[/latex].
- standard form of a quadratic function
- the function that describes a parabola, written in the form [latex]f\left(x\right)=a{\left(x-h\right)}^{2}+k[/latex], where [latex]\left(h,\text{ }k\right)[/latex] is the vertex.
- discriminant
- the value under the radical in the quadratic formula, [latex]b^2-4ac[/latex], 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
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