Explainer: Graphing Quadratic Functions

In this explainer, we will learn how to graph any quadratic function that is given in its standard and vertex forms using and studying its transformations.

Quadratic equations are used in everyday life, in science, business, and engineering. They can determine the paths of moving objects, from bouncing balls to the flight paths of bees. Business can use them to forecast revenues and design packaging to minimize waste. We can use quadratic equations to identify the minimum and maximum values of many different variables such as speed, cost, and area.

Definition: Quadratic Equation

A quadratic equation is an equation of the form 𝑦=𝑎𝑥+𝑏𝑥+𝑐, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎0.

It may also be written as 𝑦=𝑎(𝑥)+𝑘, where 𝑎 is a constant and (,𝑘) is the vertex of the turning point.

When working with functions, we may see a function written as 𝑦= or 𝑓(𝑥)=. For example, 𝑦=2𝑥3 could also be written as 𝑓(𝑥)=2𝑥3. The 𝑓(𝑥) indicates the value that would be given as an output when a particular 𝑥-value is input.

The graph of a quadratic function has a characteristic parabola shape. Depending on the value of 𝑎 in 𝑦=𝑎𝑥+𝑏𝑥+𝑐, the graph will have either a maximum or a minimum vertex (turning point). We can see how the equation 𝑦=𝑥 below has a minimum vertex with the curve opening upward and the equation with a negative coefficient of 𝑥, 𝑦=𝑥, has a maximum vertex with the curve opening downward.

A quadratic equation is a polynomial of order two, where the highest power of any of the variables is 2. If there are higher or fractional powers, then it is not a quadratic equation. So, for instance, the equation 𝑦=3𝑥+5𝑥6 is a quadratic equation in 𝑥, with values 𝑎=3, 𝑏=5, and 𝑐=6. The equation 𝑠=𝑢𝑡+12𝑎𝑡 is a quadratic in 𝑡. However, an equation such as 𝑦=5𝑥+𝑥 would not be a quadratic equation as it does not fit the form 𝑦=𝑎𝑥+𝑏𝑥+𝑐.

Let us look at some examples of how changing the coefficients 𝑎 and 𝑐 changes the appearance of the graph of a quadratic function.

We can begin by looking at the simplest example of a quadratic equation, 𝑦=𝑥, as below.

Here, the coefficient of 𝑥 is positive and the graph has a minimum vertex. We can see that the curve crosses through the point (0,0) and is symmetrical about the 𝑦-axis. We can see, in the right figure above, how changing the value of the constant 𝑐 translates the function vertically. The graph of 𝑦=𝑥+3 crosses the 𝑦-axis at (0,3), and the graph of 𝑦=𝑥1 crosses the 𝑦-axis at (0,1). A quadratic function 𝑦=𝑎𝑥+𝑏𝑥+𝑐=0 will always cross the 𝑦-axis at (0,𝑐).

Next, we will look at how changing 𝑎, the coefficient of the 𝑥-term, will change how a function appears. We can see some examples of this below with our function 𝑦=𝑥+1. Setting 𝑎=2 gives 𝑦=2𝑥+1, which dilates the graph by a factor of 2, with the vertex of the function as the center of dilation. When 𝑎=4, the function is dilated by a factor of 4, and when 𝑦=14𝑥+1, there is a dilation of factor 14 about the vertex, as we can see below. When we change the coefficient of 𝑥 in a quadratic, the vertex of the parabola is unchanged.

When 𝑎<0, the graph is inverted. Below, we can see how 𝑦=𝑥+1 and 𝑦=𝑥+1 compare.

Next, we will look at how the vertex (,𝑘) of an equation can be used to help us identify a quadratic equation in the form 𝑦=𝑎(𝑥)+𝑘. As we can see below, the function 𝑦=3(𝑥+2)+4 has a vertex of (2,4). It is important to note that because there can be many different values for 𝑎 without changing the vertex, this means that there will be many different equations that have the same vertex. We can see in the second figure below that the equations 𝑦=3(𝑥+2)+4, 𝑦=12(𝑥+2)+4, and 𝑦=3(𝑥+2)+4 all have the same vertex of (2,4). If we are using the vertex of a parabola to derive its equation, we would need to use this in conjunction with other information such as the roots of the equation, the 𝑦-intercept, and the coordinates on the curve in order to identify the correct equation.

We may see an equation written in a factored form, such as 𝑦=𝑥2𝑥24, factored as 𝑦=(𝑥+4)(𝑥6). This factored form is useful for identifying the coordinates where the equation crosses the 𝑥-axis. When a graph crosses the 𝑥-axis, we know that at these points the 𝑦-coordinate must be equal to 0. In our example equation of 𝑦=(𝑥+4)(𝑥6), we could substitute 𝑦=0 and write (𝑥+4)(𝑥6)=0.

Recall that, in any mathematical context, if we have two numbers 𝐴 and 𝐵, which multiply giving 𝐴𝐵=0, then we know that either 𝐴 or 𝐵 (or both) must equal 0.

Hence, if (𝑥+4)(𝑥6)=0, then (𝑥+4)=0(𝑥6)=0𝑥=4𝑥=6.oror

Hence, the coordinates where the graph 𝑦=(𝑥+4)(𝑥6) crosses the 𝑥-axis are (4,0) and (6,0). This process is called finding the roots of an equation.

How to Identify the Equation of a Quadratic Graph

  • Check the shape of the parabola: the graph will have a maximum turning point if 𝑎<0 and a minimum turning point if 𝑎>0.
  • Identify where the parabola crosses the 𝑦-axis: find (0,𝑐) when the equation is in the form 𝑦=𝑎𝑥+𝑏𝑥+𝑐.
  • Identify the vertex of the turning point: find (,𝑘) when the equation is in the form 𝑦=𝑎(𝑥)+𝑘.
  • Identify the roots of the equation in factored form to find the points at which the graph crosses the 𝑥-axis.
  • Recall that we can change between forms of an equation; for example, we can change 𝑦=𝑎(𝑥)+𝑘 into the form 𝑦=𝑎𝑥+𝑏𝑥+𝑐=0 by expanding the brackets and collecting terms.

Let us now look at some examples of how we can use the information shown in a graph to identify its equation.

Example 1: Identifying the Equation of a Quadratic Graph

Which of the following equations represents the graph shown?

  1. 𝑦=(𝑥3)
  2. 𝑦=𝑥(𝑥3)
  3. 𝑦=𝑥3
  4. 𝑦=𝑥+3
  5. 𝑦=(𝑥+3)


We can begin answering this by eliminating the equations that do not fit the shape of the parabola. Using the form 𝑦=𝑎𝑥+𝑏𝑥+𝑐 to find the 𝑦-intercept (0,𝑐), we can see that equation C, 𝑦=𝑥3, has a 𝑦-intercept of (0,3), which does not fit the graph. Equation D would have a 𝑦-intercept of (0,3), which also does not fit the graph.

Equation B is a factored form of the general equation 𝑦=𝑎𝑥+𝑏𝑥+𝑐, and we can expand the brackets of 𝑦=𝑥(𝑥3) to give 𝑦=𝑥3𝑥. Therefore, as the constant 𝑐 value is 0, we know that the 𝑦-intercept here would be (0,0). This does not fit the graph.

Recall that we can find the vertex (,𝑘) from the form 𝑦=𝑎(𝑥)+𝑘. This means that equation A, 𝑦=(𝑥3), has a vertex of (3,0) and equation E has a vertex of (3,0). Hence, the graph shown has the equation A, 𝑦=(𝑥3).

Example 2: Identifying the Equation of a Quadratic Graph

Which of the following graphs represents the equation 𝑦=𝑥5𝑥+8?


Since the graph of an equation 𝑦=𝑎𝑥+𝑏𝑥+𝑐=0 will cross the 𝑦-axis at (0,𝑐), we know that our equation, 𝑦=𝑥5𝑥+8, will cross the 𝑦-axis at (0,8).

As the coefficient of our 𝑥-term is a positive value of 1, this means that the graph of the function will have a minimum vertex with the curve upward.

Inspecting the graphs, we can see that the functions A and C both have these characteristics. One method of determining if an equation fits a given parabola is to select a coordinate on the parabola and check if the input, 𝑥, gives a valid output of 𝑓(𝑥).

Looking at the graph in A, we can see that the coordinate (3,2) lies on the parabola. So, for our function, 𝑓(𝑥)=𝑥5𝑥+8, if 𝑓(3)=2, then this equation would fit the parabola.

Substituting 𝑥=3 into our equation gives 𝑓(3)=35(3)+8𝑓(3)=915+8𝑓(3)=2.

Since the coordinate (3,2) lies on parabola A and does not lie on parabola C, we can give graph A as our final answer since it is the only graph with all of the properties above.

Alternatively, we could choose a coordinate on parabola C, such as (2,2), and check if 𝑓(2)=2 for the function 𝑓(𝑥)=𝑥5𝑥+8. Here, we would have 𝑓(2)=(2)5(2)+8𝑓(2)=4+10+8𝑓(2)=22.

Therefore, since 𝑓(2)2, graph C is not the graph of the function 𝑦=𝑥5𝑥+8. Graph A is the final answer.

We will now look at some examples and work through the solution to derive the equation given the shape of the parabola.

Example 3: Finding the Equation of a Quadratic Graph

Write the quadratic equation represented by the graph shown.


Recall that a quadratic can be written in the vertex form 𝑦=𝑎(𝑥)+𝑘, where the vertex is at (,𝑘). We can see that the vertex of the graph shown is at (7,0). Therefore, we can substitute =7 and 𝑘=0 into the general form of the equation, giving us 𝑦=𝑎(𝑥)+𝑘𝑦=𝑎(𝑥(7))+0𝑦=𝑎(𝑥+7).

To find the value of 𝑎 in the equation, we can select any coordinate on the parabola and substitute these 𝑥- and 𝑦-values into the equation. It is helpful to select a coordinate that has clear integer values.

We can see that the coordinate (8,1) lies on the parabola, so, substituting 𝑥=8 and 𝑦=1 into 𝑦=𝑎(𝑥+7), we have 1=𝑎(8+7)1=𝑎(1).

Since (1)=1, we have 𝑎=1, and we can substitute this into 𝑦=𝑎(𝑥+7) to give our answer for the equation: 𝑦=(𝑥+7).

Example 4: Finding the Equation of a Quadratic Graph

Write the quadratic equation represented by the graph shown. Give your answer in factored form.


In this approach, let us begin by examining the graph. We can see that the coordinates where the graph crosses the 𝑥-axis are (2,0) and (4,0), giving us the roots 2 and 4 for the equation. An equation of a function with these roots would be 𝑦=(𝑥+2)(𝑥4). However, it is important to note that there are many functions that also have these roots; for example, 𝑦=(2𝑥+4)(2𝑥8).

If we expand the brackets in the equation 𝑦=(𝑥+2)(𝑥4), we get the equation 𝑦=𝑥2𝑥8. Recall that, in this form of a quadratic equation, 𝑦=𝑎𝑥+𝑏𝑥+𝑐, we can find the 𝑦-intercept as (0,8). This would fit with the graph and is sufficient to allow us to give 𝑦=(𝑥+2)(𝑥4) as our answer.

As an alternative method, we can use the vertex of our graph to help identify its equation. Recall that a function 𝑦=𝑎(𝑥)+𝑘 has a vertex at (,𝑘). In this graph, we see that the vertex is at (1,9) which gives an equation 𝑦=𝑎(𝑥1)9. In order to find the value of 𝑎, we select a coordinate on the graph and substitute in the 𝑥- and 𝑦-values.

Selecting the coordinate (0,8), we substitute 𝑥=0 and 𝑦=8 into the equation 𝑦=𝑎(𝑥1)9, giving 8=𝑎(01)98=𝑎(1)9.

Evaluating (1) gives us 8=𝑎9.

We can now add 9 to both sides to isolate 𝑎, giving us 1=𝑎.

Therefore, 𝑎=1, and we have 𝑦=(𝑥1)9.

As we need an answer in factored form, we expand the brackets of this equation giving us 𝑦=(𝑥1)(𝑥1)9𝑦=𝑥2𝑥+19𝑦=𝑥2𝑥8.

Factoring this, we get the final answer: 𝑦=(𝑥+2)(𝑥4).

In the next example, we will look at a question regarding the roots of an equation. It can often be helpful to sketch a coordinate graph and mark key coordinates, remembering that the parabola shape of a quadratic function can be an upward or a downward curve.

Example 5: Finding the Roots of a Quadratic Graph

Given that the point (4,4) is the vertex of the graph of the quadratic function 𝑓 and 6 is a root of the equation 𝑓(𝑥)=0, find the other root of the equation.


Recall that a function of the form 𝑓(𝑥)=𝑎(𝑥)+𝑘 has the vertex (,𝑘). As we are given that the function has a vertex (4,4), we can substitute =4 and 𝑘=4 to give us