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Say you were given a quadratic function and were told to illustrate its curve in order to identify its solutions. What do you think is the best way to represent this set of solutions?

To answer your question, it is always practical to look at anything mathematical in a visual way. This allows you to look at trends and therefore examine the patterns involved. In this discussion, we shall represent quadratic functions in the form of a graph.

## Recap: Quadratic Functions in Standard Form

Before we start drawing graphs, let us first recall the definition of a quadratic equation.

The standard form of a quadratic equation is defined by the expression

$y=ax^2+bx+c$

where $$a\neq 0$$.

A quadratic expression can be set as a function by

$y=f(x)\iff f(x)=ax^2+bx+c.$

Here, $$ax^2$$ is the quadratic term,

$$bx$$ is the linear term and

$$c$$ is the constant term.

Solving quadratic equations graphically is a neat trick that enables us to determine its solutions and notice any significant behaviours present within the given expression.

## Graph of a Quadratic Function

The graph of a quadratic function has a special name as defined below.

A parabola describes the graph of any quadratic function.

Throughout this topic, we shall look at techniques for graphing such functions. Let us look at the components of a parabola.

### Components of a Parabola

Consider a general graph of $$y=ax^2+bx+c$$, as shown below.

Components of a parabola

• All parabolas have an axis of symmetry. The equation for this line is found by the formula $$x=-\frac{b}{2a}$$;

The axis of symmetry of a parabola is a vertical line that divides the parabola into two equal halves.

• The vertex is the point at which the axis of symmetry intersects a parabola:

• The $$x$$-coordinate of the vertex is $$x=-\frac{b}{2a}$$;

• The $$y$$-coordinate of the vertex is $$y=f(-\frac{b}{2a})$$;

• The $$y$$-intercept is found by plugging $$x=0$$ into the quadratic function as

\begin{align} y&=a(0)^2+b(0)+c \\ \Leftrightarrow y&= c \end{align}

• The $$x$$-intercept is found by equating the quadratic function to zero as $$ax^2+bx+c=0$$;

### Maximum and Minimum Values

The vertex of a parabola is also known as the turning point of the parabola.

The turning point of a curve is a point at which the graph changes direction, i.e., changes from increasing to decreasing or decreasing to increasing.

Looking at the image in the previous section, you can indeed confirm that a parabola changes direction and does so at the vertex.

The $$y$$-coordinate of the vertex can either be a maximum or minimum value.

The maximum value is the highest possible value of $$y$$ the curve reaches.

The minimum value is the lowest possible value of $$y$$ the curve achieves.

Considering the quadratic function $$y=ax^2+bx+c$$, the parameter which tells us beforehand if the vertex of the respective parabola will be a maximum or minimum value is the sign of the coefficient of the leading term, i.e., the sign of $$a$$. The table below describes the graph for two cases a can take, that is, $$a > 0$$ and $$a < 0$$.

 Property $$y=ax^2+bx+c$$, where $$a \neq 0$$ Value of coefficient $$a$$ $$a$$ is positive$$a > 0$$ $$a$$ is negative$$a < 0$$ Opening of Parabola Opens upward Opens downward Turning Point of $$y$$ Minimum value Maximum value Range All real numbers greater than or equal to the minimum All real numbers less than or equal to the maximum Graph Plot Example Minimum value Maximum value

In this section, we shall be introduced to three methods for graphing quadratic functions, namely

1. Making a table of values

2. Factoring method

3. Translation method

Each technique will include a step-by-step method followed by a worked example so that you will become familiarised with its procedure.

### Making a Table of Values

For this technique of solving quadratic equations, we aim to graph such expressions by making a table of values. The steps for this method are explained below.

1. Set $$y=ax^2+bx+c$$;

2. Find the $$y$$-intercepts by plugging $$x = 0$$ into $$y$$;

3. Locate the axis of symmetry and vertex;

4. Plug-in several values for $$x$$ into $$y$$ and create a table of values for this calculated set of points;

5. Determine the $$x$$-intercepts. If the solutions cannot be found explicitly, we can estimate the solutions by identifying the integers between which the solutions are located;

6. Plot the graph.

In Step 5, we shall use the Location Principle to estimate the location of the solutions on a given quadratic equation. Below is an explanation of the Location Principle followed by three worked examples.

#### The Location Principle

Suppose $$y = f(x)$$ represents a quadratic function and a and b are two real numbers such that $$f(a) < 0$$ and $$f(b) > 0$$. Then f has at least one real zero between $$a$$ and $$b$$.

How does the Location Principle work?

The $$x$$-intercept of a graph crosses the $$x$$-axis at $$y = 0$$. This, in turn, changes the sign of the $$y$$-values that follow in the graph.

Essentially, the Location Principle tells us that we are looking for a change in sign between two outputs of $$y$$ given two inputs of $$x$$.

Now, to an example.

Consider the function $f(x)=x^2-6x+3.$
1. Determine the $$y$$-intercept, the equation of the axis of symmetry, and the $$x$$-coordinate of the vertex.
2. Create a table of values that includes the vertex.
3. Graph the function based on the results from Questions 1 and 2.

Solution

Here, $$a = 1$$, $$b = -6$$ and $$c = 3$$.

1. To find the $$y$$-intercept we shall evaluate $$f$$ at $$x = 0$$. In doing so we obtain, $$f(0) = c = 3$$. Thus the $$y$$-intercept is $$(0, 3)$$. The equation of the axis of symmetry is found by using the formula

$x=-\frac{(-6)}{2(1)}\implies x=3$

Therefore, the equation of the axis of symmetry is $$x = 3$$. It follows that the $$x$$-coordinate of the vertex is 3.

2. We shall now make our table of values. To do this, choose some values for $$x$$ that are less than 3 and some that are greater than 3. This ensures that the points on each side of the axis of symmetry are graphed.

 $$x$$ 0 1 2 3 4 5 6 $$f(x)$$ 3 –2 –5 –6 –5 –2 3

Let us apply the Location Principle to this example. Looking at the table above, there happens to be a change in sign between $$x = 0$$ and $$x = 1$$. This indicates that there is one real zero between these two values of $$x$$ (point A). Similarly, we observe a change in sign between $$x = 5$$ and $$x = 6$$ (point B) which again tells us that one real zero must exist between this pair of $$x$$ values.

3. The graph is shown below.

Graph of Example 1

### Factoring Method

We can factor quadratic expressions by identifying certain patterns in a given expression. Factoring is a method where we simplify the standard form of a quadratic equation into the intercept form. This will allow us to locate the roots of the expression and plot the $$x$$-intercepts for the corresponding graph. There are four ways in which we can factor quadratic equations. This is described in the table below.

 Factoring Technique General Case Greatest Common Factor (GCF) $a^3b^2-3ab^2=ab^2(a^2-3)$ Difference of Two Squares $a^2-b^2=(a+b)(a-b)$ Perfect Square Trinomials $a^2+2ab+b^2=(a+b)^2$$a^2-2ab+b^2=(a-b)^2$ General Trinomials $acx^2+(ad+bc)x+bd=(ax+b)(cx+d)$

$6x^2=1-x$

and graph the respective function.

Solution

We begin by rearranging the equation to the standard form of a quadratic equation

$6x^2+x-1=0$

This is a general trinomial in which we can factorise by $$(ax+b)(cx+d)$$. Note that the coefficient 6 can be factorised as $$6\times 1$$ or $$2\times 3$$ while the constant $$-1$$ can be factorised as $$-1\times 1$$.

Remember that you have to test all possible pairs and positions for these two products. Using trial and error, we can factorise this as

$(3x-1)(2x+1)=0$

By the Zero Product Property, the roots of this function are

$3x-1=0\implies x=\frac{1}{3}$

and

$2x+1=0\implies x=-\frac{1}{2}$

Since the coefficient of $$x^2$$ is $$a = 6 > 0$$, the parabola opens upward. The first root is indicated by the letter A and the second by B. The graph is shown below.

Graph of Example 2

### Translation Method and the Vertex Form

In this section, we shall be introduced to the vertex form of a quadratic function. Here, we shall refer to the standard quadratic equation, that is, $$y=x^2$$ and observe how manipulating certain coefficients can change the shape of this simple graph. Let us begin with the following definition.

The vertex form of a quadratic equation is described by the expression

$y=a(x-h)^2+k,$

where $$(h, k)$$ is the vertex of the parabola and $$x=h$$ is the axis of symmetry.

The vertex form is a clever way to express quadratic functions so that we can translate the graph accordingly based on the standard quadratic function. First, let us recall what a translation is.

A translation is a geometrical transformation that slides a figure without changing its shape or size.

We can apply this concept in graphing quadratic functions by altering the coefficients or constants in a given expression. The graph of the basic quadratic function is $$f(x)=x^2$$. This is shown below.

The vertex here is the origin, $$(0, 0)$$ and the axis of symmetry is $$x=0$$.

Say we are given a quadratic equation in vertex form. There are three ways in which we can transform this graph. This is illustrated in the table below.

 Change of Variable Affect of Change Plot of Graph Varying $$a$$ alters the function in the $$y$$-direction, (the coefficient $$a$$ affects the steepness of the graph) If $$a$$ is large, the graph becomes vertically stretched (red line)If $$a$$ is small, the graph vertically elongated (blue line) If $$a$$ is negative, the graph becomes inverted (green line) Varying variable $$a$$ Varying $$h$$ changes the function along the $$x$$-axis by $$h$$ units If $$h$$ is negative, the graph shifts $$h$$ units to the left of the $$x$$-axis (red line)If $$h$$ is positive, the graph shifts $$h$$ units to the right of the $$x$$-axis (blue line) Varying variable $$h$$ Varying $$k$$ translates the function up or down the $$y$$-axis by $$k$$ units If $$k$$ is negative, the graph moves down $$k$$ units in the $$y$$-axis (red line)If $$k$$ is positive, the graph moves up $$k$$ units in the $$y$$-axis (blue line) Varying variable $$k$$

Express the equation

$y=x^2+4x+6$

in vertex form and plot the graph of the function.

Solution

To translate this function, we first need to rewrite this function in its corresponding vertex form. We shall conduct the method of completing the square to achieve this. We will write the right-hand side of the expression as

$y=(x^2+4x+?)+6-?$

To find the missing values, we add $$(\frac{4}{2})^2=4$$ and balance the equation by subtracting 4 as

$y=(x^2+4x+4)+6-4$

Now writing this as a perfect square, we obtain

$y=(x+2)^2+2$

Since $$h = –2$$ and $$k = 2$$, the vertex is at $$(–2, 2)$$. The axis of symmetry is $$x = –2$$. Since $$a = 1$$, the graph opens up and has the same shape as the graph of $$y=x^2$$. Furthermore, the graph is translated 2 units left and 2 units up.

The graph is shown below.

Graph of Example 3

## Examples of Quadratic Function Graphs

In this final segment, we shall look at several more examples of graphing quadratic functions. Here, we will gather everything we have discussed throughout this article to answer the following problems.

$x^2-25=0$

and graph the respective function.

Solution

Notice that $$25=5^2$$. Thus, the equation takes the form of the difference between two squares. Therefore, we can factorize this as

$$x^2-25=0\implies (x-5)(x+5)=0$$

By the Zero Product Property, the roots of this function are

$x-5=0\implies x=5$

and

$x+5=0\implies x=5$

Since the coefficient of $$x^2$$ is a = 1 > 0, the parabola opens upward. The first root is indicated by A and the second by B. The graph is shown below.

Graph of Example 4

Let us look at another example.

Consider the function

$f(x)=2x^2-4x-3$

1. Decide whether the function has a maximum or minimum value.
2. Evaluate the maximum or minimum value of the function.
3. State the domain and range of the function.

Solution

Here, $$a = 2$$, $$b = –4$$, and $$c = –3$$.

1. Since $$a > 0$$, the graph opens up and the function has a minimum value.

2. The minimum value of the function is the $$y$$-coordinate of the vertex.

The $$x$$-coordinate of the vertex is

$-\frac{-4}{2(2)}=1$

To find the y-coordinate of the vertex, we shall evaluate the function at $$x = 1$$.

$f(1)=2(1)^2-4(1)-3\implies f(1)=-5$

Hence, the minimum value of the function is –5.

3. The domain is the set of all real numbers. The range is the set of all real numbers greater than or equal to the minimum value, $$f(x)\geq -5$$. In set notation, $$[–5, +∞ ]$$.

The graph is shown below.

Graph of Example 5

Write the equation

$y=-2x^2-12x+17$

in vertex form and plot the graph of the function.

Solution

We begin by factoring the first two terms of the right-hand side as

$y=-2(x^2+6x)+17$

Next, we shall complete the square of this expression.

$y=-2(x^2+6x+?)+17-(-2)?$

To find the missing values, we add $$(\frac{6}{2})^2=9$$ and balance the equation by subtracting –2(9) as

$y=-2(x^2+6x+9)+17-(-2)(9)$

Now writing this as a perfect square, we obtain

$y=-2(x+3)^2+35$

The vertex is at $$(–3, 35)$$, and the axis of symmetry is $$x = –3$$. Since $$a = –2$$, the graph opens downward and is narrower than the graph of $$y=x^2$$. Additionally, the graph is translated 1 unit right and 2 units up.

The graph is shown below.

Graph of Example 6

## Quadratic Function Graphs - Key takeaways

• The standard form of a quadratic function is $$ax^2+bx+c=0$$ where $$a \neq 0$$.
• The $$y$$-intercept is found by plugging $$x = 0$$ into the quadratic function.
• The $$x$$-intercept is found by equating the quadratic function to zero, $$f(x) = 0$$.
• If $$a > 0$$, the graph has a minimum point and the parabola opens upward.
• If $$a < 0$$, the graph has a maximum point and the parabola opens downward.
• We can use translations to graph a quadratic function.
• For the vertex form $$y=a(x-h)^2+k$$
• Varying $$a$$ alters the function in the $$y$$-direction
• Varying $$h$$ changes the function along the $$x$$-axis by $$h$$ units
• Varying $$k$$ translates the function up or down the $$y$$-axis by $$k$$ units

1. Set f(x) = ax2 + bx + c;

2. Find the y-intercepts by plugging x = 0 into f(x);

3. Locate the axis of symmetry and vertex;

4. Plug-in several values for x into f(x) and create a table of values for this calculated set of points;

5.  Determine the x-intercepts. If the solutions cannot be found explicitly, we can estimate the solutions by identifying the integers between which the solutions are located;

6. Plot the graph.

The curve is in the shape of a parabola with one turning point.

The graph of a quadratic function is called a parabola.

You need to find the x-intercept, y-intercept, vertex and determine the shape of the parabola based on the coefficient of x2

The graph of a quadratic function forms a parabola.

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