Quadratic Functions

Curved antennas 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.

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

Recognizing 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. These features are illustrated below.

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

Solution. 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.

Understanding how the graphs of parabolas are related to their quadratic functions

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

where $a$, $b$, and $c$ are real numbers and $a\ne0$. 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 $x=-\tfrac{b}{2a}$. If we use the quadratic formula, $x=\tfrac{-b\pm\sqrt{b^2-4ac}}{2a}$, to solve $ax^2+bx+c=0$ for the $x$-intercepts, or zeros, we find the value of $x$ halfway between them is always $x=-\tfrac{b}{2a}$, the equation for the axis of symmetry.

The graph below is the quadratic function written in general form as $y=x^2+4x+3$. In this form, $a=1$, $b=4$, and $c=3$. Because $a>0$, the parabola opens upward. The axis of symmetry is $x=-\tfrac{4}{2(1)}=-2$. 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, $(-2,-1)$. The $x$-intercepts, those points where the parabola crosses the $x$-axis, occur at $(-3,0)$ and $(-1,0)$.

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

where $(h,k)$ 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.

As with the general form, if $a>0$, the parabola opens upward and the vertex is a minimum. If $a<0$, the parabola opens downward, and the vertex is a maximum. The graph below is the quadratic function written in standard form as $y=-3(x+2)^2+4$. Since $x-h=x+2$ in this example, $h=-2$. In this form, $a=-3$, $h=-2$, and $k=4$. Because $a<0$, the parabola opens downward. The vertex is at $(-2,4)$.

The standard form is useful for determining how the graph is transformed from the graph of $y=x^2$. The graph below is that basic function.

If $k>0$, the graph shifts upward, whereas if $k<0$, the graph shifts downward. Above, $k>0$, so the graph is shifted 4 units upward. If $h>0$, the graph shifts toward the right and if $h<0$, the graph shifts to the left. Above, $h<0$, so the graph is shifted 2 units to the left. The magnitude of $a$ indicates the stretch of the graph. If $|a|>1$, the point associated with a particular $x$-value shifts farther from the $x$-axis, so the graph appears to become narrower, and there is a vertical stretch. But if $|a|<1$, the point associated with a particular $x$-value shifts closer to the $x$-axis, so the graph appears to become wider, but in fact there is a vertical compression. Above, $|a|>1$, so the graph becomes narrower.

The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.

For the linear terms to be equal, the coefficients must be equal.

This is the axis of symmetry we defined earlier. Setting the constant terms equal:

In practice, though, it is usually easier to remember that $k$ is the output value of the function when the input is $h$, so $f(h)=k$.

Example. Write an equation for the quadratic function $g$ shown below as a transformation of $f(x)=x^2$, and then expand the formula, and simplify terms to write the equation in general form.

Solution. We can see the graph of $g$ is the graph of $f(x)=x^2$ shifted to the left 2 and down 3, giving a formula in the form $g(x)=a(x+2)^2-3$.

Substituting the coordinates of a point on the curve, such as $(0,-1)$, we can solve for the stretch factor.

In standard form, the algebraic model for this graph is $g(x)=\tfrac{1}{2}(x+2)^2-3$.

To write this in general polynomial form, we can expand the formula and simplify terms.

Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.

Analysis. We can check our work using the table feature on a graphing utility. First enter $\text{Y1}=\tfrac{1}{2}(x+2)^2-3$. Next, select $\text{TBLSET}$, then use $\text{TblStart}=-6$ and $\Delta\text{Tbl}=2$, and select $\text{TABLE}$.

The ordered pairs in the table correspond to points on the graph.

A coordinate grid can be superimposed over the quadratic path of a basketball, with the shooter releasing the ball at the origin. Assume the point $(-4,7)$ is the highest point of the basketball’s trajectory, and that the hoop sits 4 feet high at the horizontal position $x=-7.5$.

Example. Find the vertex of the quadratic function $f(x)=2x^2-6x+7$. Rewrite the quadratic in standard form (vertex form).

Solution. The horizontal coordinate of the vertex will be at

The vertical coordinate of the vertex will be at

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

Using the vertex to determine the shifts,

Analysis. 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, $(x)$.

Finding the domain and range of a quadratic function

Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers. Because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all $y$-values greater than or equal to the $y$-coordinate at the turning point or less than or equal to the $y$-coordinate at the turning point, depending on whether the parabola opens up or down.

Example. Find the domain and range of $f(x)=-5x^2+9x-1$.

Solution. As with any quadratic function, the domain is all real numbers.

Because $a$ is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the $x$-value of the vertex.

The maximum value is given by $f(h)$.

The range is $f(x)\le\tfrac{61}{20}$, or $\left(-\infty,\tfrac{61}{20}\right]$.

Determining the maximum and minimum values of quadratic functions

The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola. We can see the maximum and minimum values in the two graphs below.

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. 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.

1. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length $L$.
2. What dimensions should she make her garden to maximize the enclosed area?

Solution. Let’s use a diagram such as the one below to record the given information. It is also helpful to introduce a temporary variable, $W$, to represent the width of the garden and the length of the fence section parallel to the backyard fence.

1. We know we have only 80 feet of fence available, and $L+W+L=80$, or more simply, $2L+W=80$. This allows us to represent the width, $W$, in terms of $L$.

   $$W=80-2L$$

   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

   $$\begin{array}{rclcl} A &=& LW &=& L(80-2L) \\[4pt] A(L) &=& 80L-2L^2 && \end{array}$$

   This formula represents the area of the fence in terms of the variable length $L$. The function, written in general form, is

   $$A(L)=-2L^2+80L$$

2. 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 $a$ is the coefficient of the squared term, $a=-2$, $b=80$, and $c=0$.

   To find the vertex:

   $$\begin{array}{lrcl} & h &=& -\tfrac{80}{2(-2)} \\[4pt] & &=& 20 \end{array}$$

   and

   $$\begin{array}{lrcl} & k &=& A(20) \\[4pt] & &=& 80(20)-2(20)^2 \\[4pt] & &=& 800 \end{array}$$

   The maximum value of the function is an area of 800 square feet, which occurs when $L=20$ 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. This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on the graph below.

Example. 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?

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, $p$ for price per subscription and $Q$ for quantity, giving us the equation $\text{Revenue}=pQ$.

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently $p=30$ and $Q=84{,}000$. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, $p=32$ and $Q=79{,}000$. 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 $y$-intercept.

This gives us the linear equation $Q=-2{,}500p+159{,}000$ 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.

Analysis. This could also be solved by graphing the quadratic, below. We can see the maximum revenue on the graph of the quadratic function.

Finding the x- and y-intercepts of a quadratic function

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 below that the number of $x$-intercepts can vary depending upon the location of the graph.

Example. Find the $y$- and $x$-intercepts of the quadratic $f(x)=3x^2+5x-2$.

Solution. We find the $y$-intercept by evaluating $f(0)$.

So the $y$-intercept is at $(0,-2)$.

For the $x$-intercepts, we find all solutions of $f(x)=0$.

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

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

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

Rewriting quadratics in standard form

In the example above, 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.

Example. Find the $x$-intercepts of the quadratic function $f(x)=2x^2+4x-4$.

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

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

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

So now we can rewrite in standard form.

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

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

Analysis. We can check our work by graphing the given function on a graphing utility and observing the $x$-intercepts, shown below.

Example. Solve $x^2+x+2=0$.

Solution. Let’s begin by writing the quadratic formula: $x=\tfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.

When applying the quadratic formula, we identify the coefficients $a$, $b$, and $c$. For the equation $x^2+x+2=0$, we have $a=1$, $b=1$, and $c=2$. Substituting these values into the formula we have:

The solutions to the equation are $\tfrac{-1+i\sqrt{7}}{2}$ and $\tfrac{-1-i\sqrt{7}}{2}$, or $-\tfrac{1}{2}+\tfrac{\sqrt{7}}{2}i$ and $-\tfrac{1}{2}-\tfrac{\sqrt{7}}{2}i$.

Example. 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 $H(t)=-16t^2+80t+40$.

1. When does the ball reach the maximum height?
2. What is the maximum height of the ball?
3. When does the ball hit the ground?

Solution.

1. The ball reaches the maximum height at the vertex of the parabola.

   $$\begin{array}{lrcl} & h &=& -\tfrac{80}{2(-16)} \\[4pt] & &=& \tfrac{80}{32} \\[4pt] & &=& \tfrac{5}{2} \\[4pt] & &=& 2.5 \end{array}$$

   The ball reaches a maximum height after 2.5 seconds.

2. To find the maximum height, find the $y$-coordinate of the vertex of the parabola.

   $$\begin{array}{lrcl} & k &=& H\!\left(-\tfrac{b}{2a}\right) \\[4pt] & &=& H(2.5) \\[4pt] & &=& -16(2.5)^2+80(2.5)+40 \\[4pt] & &=& 140 \end{array}$$

   The ball reaches a maximum height of 140 feet.

3. To find when the ball hits the ground, we need to determine when the height is zero, $H(t)=0$.

   We use the quadratic formula.

   $$\begin{array}{lrcl} & t &=& \tfrac{-80\pm\sqrt{80^2-4(-16)(40)}}{2(-16)} \\[4pt] & &=& \tfrac{-80\pm\sqrt{8{,}960}}{-32} \end{array}$$

   Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

   $$t=\tfrac{-80-\sqrt{8{,}960}}{-32}\approx5.458 \qquad\text{or}\qquad t=\tfrac{-80+\sqrt{8{,}960}}{-32}\approx-0.458$$

   The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See the graph below.

Key equations

Key concepts

• A polynomial function of degree two is called a quadratic function.
• The graph of a quadratic function is a parabola. A parabola is a U-shaped curve that can open either up or down.
• The axis of symmetry is the vertical line passing through the vertex. The zeros, or $x$-intercepts, are the points at which the parabola crosses the $x$-axis. The $y$-intercept is the point at which the parabola crosses the $y$-axis.
• Quadratic functions are often written in general form. Standard or vertex form is useful to easily identify the vertex of a parabola. Either form can be written from a graph.
• The vertex can be found from an equation representing a quadratic function.
• The domain of a quadratic function is all real numbers. The range varies with the function.
• A quadratic function’s minimum or maximum value is given by the $y$-value of the vertex.
• The minimum 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.

Practice

Recognize characteristics of parabolas

Understand how the graph of a parabola is related to its quadratic function

Determine a quadratic function’s minimum or maximum value

Solve problems involving a quadratic function’s minimum or maximum value

A backyard farmer wants to enclose a rectangular corral using 200 feet of fencing on all four sides.

This section is adapted from Precalculus 2e, Section 3.2: Quadratic Functions by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated every graph as an accessible inline SVG generated from its exact equation — the labeled-features schematic parabola $y=x^2-2x-3$; the vertex/no-x-intercept parabola $y=\tfrac{2}{3}(x-3)^2+1$ of Example 1; the general-form parabola $y=x^2+4x+3$; the standard-form parabola $y=-3(x+2)^2+4$; the basic parabola $y=x^2$; the graph-transformation parabola $g(x)=\tfrac{1}{2}(x+2)^2-3$; the two minimum/maximum panels $f(x)=(x-2)^2+1$ and $g(x)=-(x+3)^2+4$; the area function $A(L)=-2L^2+80L$; the revenue function, scaled to thousands of dollars, $-2.5p^2+159p$; the three no/one/two-$x$-intercept panels; the $y$-/$x$-intercept parabola $f(x)=3x^2+5x-2$; the standard-form parabola $f(x)=2x^2+4x-4$ with its irrational intercepts; the corrected height-vs-time parabola $H(t)=-16t^2+80t+40$; and the two Practice graph-reading items $f(x)=x^2-4x+1$ and $f(x)=-2x^2+8x-1$; recreated the garden diagram as a labeled rectangle-in-rectangle figure; presented the table-feature check in Example 2 as a Markdown table; omitted the decorative satellite-dish photograph, which carries no mathematics, and reworded the opening sentence that pointed at it; omitted the section’s four Media links to external graphing-calculator resources; converted the basketball Try It’s photograph-with-grid-overlay into a recreated graph, stating in the question the release point, vertex, and the hoop’s location and height that the source conveyed only through the image; converted the practice problems (“Try Its”) into interactive exercises with instant feedback — a fillin plus multiple choice for the basketball path and shot outcome, a multiple choice among competing writings plus a fillin for the rewritten-in-standard-form Try It (its general-form half is a bare reordering of an already-expanded printed expression, so no answerForm token restricts term order and a retyped printed span would grade correct; a multiple choice among the general, standard, and mis-ordered forms grades the shape instead), a fillin for the domain-and-range Try It, a fillin plus multiple choice for the $y$-intercept and $x$-intercept count of the rewritten function, and three fillins for the falling-rock application; split each of the two “sketch a graph and give the vertex, axis of symmetry, and intercepts” Practice items into a graphing or fillin component per requested quantity; and adapted thirteen selected end-of-section exercises — two vertex/axis-of-symmetry rewrites, two graph-reading equations, two minimum-value/axis-of-symmetry evaluations, a range evaluation, and four real-world optimization problems (a four-sided corral, a ticket-revenue maximization, a rocket’s maximum height, and a smallest-product pair) — into nineteen interactive components in a closing Practice block, one group per objective. One upstream figure defect is corrected here: the pinned CNXML’s Figure 16 (module m49337) prints its vertical axis labeled “$t$” carrying the height values (50, 100, 150) and its horizontal axis labeled “$H$” carrying the time values (1 through 6), the reverse of the function $H(t)$ it illustrates, alongside a stray “$(x$ from $-1$ to $6)$” annotation left over from a different window setting; this page draws the corrected axes, time on the horizontal axis and height on the vertical axis, over the function’s own domain, and carries a visible source note beside the corrected figure disclosing the correction in addition to this footer.