Use the quadratic formula and factoring to find both real and complex roots ([latex]x[/latex]-intercepts) of quadratic functions.
Use algebra to find the [latex]y[/latex]-intercepts of a quadratic function.
Solve problems involving the roots and intercepts of a quadratic function.
Use the discriminant to determine the nature (real or complex) and quantity of solutions to quadratic equations.
Determine a quadratic function’s minimum or maximum value.
Solve problems involving a quadratic function’s minimum or maximum value.
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.
Intercepts of Quadratic Functions
We also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the [latex]y[/latex]-intercept of a quadratic by evaluating the function at an input of zero, and we find the [latex]x[/latex]-intercepts at locations where the output is zero. Notice that the number of [latex]x[/latex]-intercepts can vary depending upon the location of the graph.
Number of [latex]x[/latex]-intercepts of a parabola
Mathematicians also define [latex]x[/latex]-intercepts as roots of the quadratic function.
Example: Solving a Quadratic Equation with the Quadratic Formula
Solve [latex]{x}^{2}+x+2=0[/latex].
Show Solution
Let’s begin by writing the quadratic formula: [latex]x=\dfrac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex].
When applying the quadratic formula, we identify the coefficients [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex]. For the equation [latex]{x}^{2}+x+2=0[/latex], we have [latex]a=1[/latex], [latex]b=1[/latex], and [latex]c=2[/latex]. Substituting these values into the formula we have:
The solutions to the equation are [latex]x=\dfrac{-1+i\sqrt{7}}{2}[/latex] and [latex]x=\dfrac{-1-i\sqrt{7}}{2}[/latex] or [latex]x=-\dfrac{1}{2}+\dfrac{i\sqrt{7}}{2}[/latex] and [latex]x=-\dfrac{1}{2}-\dfrac{i\sqrt{7}}{2}[/latex].
Analysis of the Solution
This quadratic equation has only non-real solutions.
How To: Given a quadratic function, find the x-intercepts by rewriting in standard form.
Substitute [latex]a[/latex] and [latex]b[/latex] into [latex]h=-\dfrac{b}{2a}[/latex].
Substitute [latex]x=h[/latex] into the general form of the quadratic function to find [latex]k[/latex].
Rewrite the quadratic in standard form using [latex]h[/latex] and [latex]k[/latex].
Solve for when the output of the function will be zero to find the [latex]x[/latex]–intercepts.
Example: Finding the Roots of a Parabola
Find the [latex]x[/latex]-intercepts of the quadratic function [latex]f\left(x\right)=2{x}^{2}+4x - 4[/latex].
Show Solution
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.
The graph has [latex]x[/latex]–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.
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Example: Applying the Vertex and x-Intercepts of a Parabola
A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation [latex]H\left(t\right)=-16{t}^{2}+80t+40[/latex].
a. When does the ball reach the maximum height?
b. What is the maximum height of the ball?
c. When does the ball hit the ground?
Show Solution
a. The ball reaches the maximum height at the vertex of the parabola.
Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.
[latex]t=\dfrac{-80+\sqrt{8960}}{-32}\approx 5.458\hspace{3mm}[/latex] or [latex]\hspace{3mm}t=\dfrac{-80-\sqrt{8960}}{-32}\approx -0.458[/latex]
Since the domain starts at [latex]t=0[/latex] when the ball is thrown, the second answer is outside the reasonable domain of our model. The ball will hit the ground after about 5.46 seconds.
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A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation [latex]H\left(t\right)=-16{t}^{2}+96t+112[/latex].
a. When does the rock reach the maximum height?
b. What is the maximum height of the rock?
c. When does the rock hit the ocean?
Show Solution
a. 3 seconds b. 256 feet c. 7 seconds
Complex Roots
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 [latex]x[/latex]-axis, therefore it doesn’t have any [latex]x[/latex]-intercepts. Recall that the [latex]x[/latex]-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 [latex]x[/latex]-intercepts because the function does not contain [latex](x,y)[/latex] pairs that are on the [latex]x[/latex]-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 are complex numbers.
Example
Find the [latex]x[/latex]-intercepts of the quadratic function. [latex]f(x)=x^2+2x+3[/latex]
Show Solution
The [latex]x[/latex]-intercepts of the function [latex]f(x)=x^2+2x+3[/latex] are found by setting it equal to zero, and solving for [latex]x[/latex] since the [latex]y[/latex] values of the [latex]x[/latex]-intercepts are zero.
First, identify [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex].
[latex]x^2+2x+3=0[/latex]
[latex]a=1,b=2,c=3[/latex]
Substitute these values into the quadratic formula.
The solutions to this equation are complex, therefore there are no [latex]x[/latex]-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 [latex]x[/latex]-intercepts in the real numbers.
Note how the graph does not cross the [latex]x[/latex]-axis, therefore there are no real [latex]x[/latex]-intercepts for this function.
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The following video gives another example of how to use the quadratic formula to find complex solutions to a quadratic equation.
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]
Show Solution
Calculate the discriminant [latex]{b}^{2}-4ac[/latex] for each equation and state the expected type of solutions.
[latex]\begin{align}{x}^{2}+4x+4=0&&{b}^{2}-4ac={\left(4\right)}^{2}-4\left(1\right)\left(4\right)=0\end{align}\hspace{4mm}[/latex]There will be one repeated rational solution.
[latex]\begin{align}8{x}^{2}+14x+3=0&&{b}^{2}-4ac={\left(14\right)}^{2}-4\left(8\right)\left(3\right)=100\end{align}\hspace{4mm}[/latex]As [latex]100[/latex] is a perfect square, there will be two rational solutions.
[latex]\begin{align}3{x}^{2}-5x - 2=0&&{b}^{2}-4ac={\left(-5\right)}^{2}-4\left(3\right)\left(-2\right)=49\end{align}\hspace{4mm}[/latex]As [latex]49[/latex] is a perfect square, there will be two rational solutions.
[latex]\begin{align}3{x}^{2}-10x+15=0&&{b}^{2}-4ac={\left(-10\right)}^{2}-4\left(3\right)\left(15\right)=-80\end{align}\hspace{4mm}[/latex]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 [latex]\sqrt{{{b}^{2}}-4ac}[/latex], and how it helps to determine the solution set.
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 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[/latex]” 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.
Show Solution
Evaluate [latex]b^{2}-4ac[/latex]. First note that [latex]a=1,b=−4[/latex], and [latex]c=10[/latex].
The result is a negative number. The discriminant is negative, so [latex]x^{2}-4x+10=0[/latex] has two complex solutions.
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Applications With Quadratic Functions
There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.
Example: Finding the Maximum Value of a Quadratic Function
A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.
Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[/latex].
What dimensions should she make her garden to maximize the enclosed area?
Show Solution
Let’s use a diagram such as the one above to record the given information. It is also helpful to introduce a temporary variable, [latex]W[/latex], to represent the width of the garden and the length of the fence section parallel to the backyard fence.
We know we have only 80 feet of fence available, and [latex]L+W+L=80[/latex], or more simply, [latex]2L+W=80[/latex]. This allows us to represent the width, [latex]W[/latex], in terms of [latex]L[/latex].
[latex]W=80 - 2L[/latex]Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so [latex]A=LW=L\left(80 - 2L\right)=80L - 2{L}^{2}[/latex]. This formula represents the area of the fence in terms of the variable length [latex]L[/latex]. The function, written in general form, is[latex]A\left(L\right)=-2{L}^{2}+80L[/latex].
The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since [latex]a[/latex] is the coefficient of the squared term, [latex]a=-2,b=80[/latex], and [latex]c=0[/latex].
The maximum value of the function is an area of 800 square feet, which occurs when [latex]L=20[/latex] feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.
Analysis of the Solution
This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function below.
The problem we solved above is called a constrained optimization problem. We can optimize our desired outcome given a constraint, which in this case was a limited amount of fencing materials. Try it yourself in the next problem.
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How To: Given an application involving revenue, use a quadratic equation to find the maximum.
Write a quadratic equation for revenue.
Find the vertex of the quadratic equation.
Determine the [latex]y[/latex]-value of the vertex.
Example: Finding Maximum Revenue
The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?
Show Solution
Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, [latex]p[/latex] for price per subscription and [latex]Q[/latex] for quantity, giving us the equation [latex]\text{Revenue}=pQ[/latex].
Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently [latex]p=30[/latex] and [latex]Q=84,000[/latex]. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, [latex]p=32[/latex] and [latex]Q=79,000[/latex]. From this we can find a linear equation relating the two quantities. The slope will be
This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the [latex]y[/latex]-intercept.
[latex]\begin{align}&Q=-2500p+b &&\text{Substitute in the point }Q=84,000\text{ and }p=30 \\ &84,000=-2500\left(30\right)+b &&\text{Solve for }b \\ &b=159,000 \end{align}[/latex]
This gives us the linear equation [latex]Q=-2,500p+159,000[/latex] relating cost and subscribers. We now return to our revenue equation.
We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.
The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.
This could also be solved by graphing the quadratic. We can see the maximum revenue on a graph of the quadratic function.
In the example above, we knew the number of subscribers to a newspaper and used that information to find the optimal price for each subscription. What if the price of subscriptions is affected by competition?
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Previously, we found a quadratic function that modeled revenue as a function of price.
We found that selling the paper at [latex]\$31.80[/latex] per subscription would maximize revenue. What if your closest competitor sells their paper for [latex]\$25.00[/latex] per subscription? What is the maximum revenue you can make you sell your paper for the same?
Show Solution
Evaluating the function for [latex]p=25[/latex] gives [latex]\$2,412,500[/latex].
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Key Equations
The quadratic formula [latex]x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}[/latex]
The discriminant is defined as [latex]b^2-4ac[/latex]
Key Concepts
The zeros, or [latex]x[/latex]-intercepts, are the points at which the parabola crosses the [latex]x[/latex]-axis. The [latex]y[/latex]-intercept is the point at which the parabola crosses the [latex]y[/latex]–axis.
The vertex can be found from an equation representing a quadratic function.
A quadratic function’s minimum or maximum value is given by the [latex]y[/latex]-value of the vertex.
The minium or maximum value of a quadratic function can be used to determine the range of the function and to solve many kinds of real-world problems, including problems involving area and revenue.
Some quadratic equations must be solved by using the quadratic formula.
The vertex and the intercepts can be identified and interpreted to solve real-world problems.
Some quadratic functions have complex roots.
Glossary
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
vertex form of a quadratic function
another name for the standard form of a quadratic function
zeros
in a given function, the values of [latex]x[/latex] at which [latex]y=0[/latex], also called roots
