Have you ever thrown a ball in the air and watched the way it moves? The up and down motion of the throw takes the form of a parabola - the same form a quadratic graph takes.
This means you could have a quadratic function that models the throw and trajectory of the ball and the graph of this function would model the motion.
This article will explore and explain the standard form of a quadratic function and determine the effect of varying coefficients on the graph of this polynomial. It will also explore inequalities in the form of quadratic functions.
What are Quadratic Graphs?
A quadratic graph is a parabola plotted based on a polynomial of order two, that is, a polynomial where the highest power of \(x\) is 2.
In figure 1 you can see two quadratic graphs. As you can see in the labels, the highest power in the polynomial is that of \(x^2\). Read on more to find out why this is.
Fig. 1 - Types of quadratic polynomials and their graphs.
In order to understand quadratic graphs, we'll introduce the Standard form of a quadratic function.
Standard form of a quadratic function
The standard form of a quadratic function (or a polynomial of order 2) is given by \[f(x)=a(x-h)^2+k,\] where \(a,h\) and \(k\) are real numbers.
Tweaking any of these constants will have a visible effect on the graph of the function. The reason for which this version is utilized in graphing over the general form \[f(x)=ax^2+bx+c\] is that, although the constant \(a\) has the same effect in both forms,
in standard form, shifting any of the constants \(h\) or \(k\) has a one-dimensional effect, i.e., the graph moves to the right or to the left, up or down (as you'll see further on the article);
whereas shifting the constant \(b\) in the general form has a two-dimensional effect, i.e., the graph moves up and right or down and left, thus complicating manually graphing this form.
Now that you know more about the standard form of quadratic functions, let's move to see how this form can help us graph quadratics.
Graphing quadratics in standard form
Before going through the graphs of quadratic functions in standard form, you'll learn more about crucial concepts that are essential to graph our quadratics perfectly.
Let's start with the Opening of a parabola.
Opening of a parabola
With reference to the standard form of a quadratic function, we will now explore the effects of varying each of the constants, \(a\), \(h\) and \(k\), individually.
The constant \(a\) has two effects on the parabola of a quadratic function, the first of which is dilation, that is, expanding or contracting the parabola according to its absolute value. You'll see more on this in the section Dilation.
The second effect is on defining the direction of the opening of the parabola according to its sign:
The effects that the sign of \(a\) has on the opening of a parabola can be seen in figure 2.
Fig. 2 - Effects of the sign of \(a\). Left: positive \(a\) opens upward. Right: Negative \(a\) opens downward.
Notice that a value of \(a\) of \(0\) results in a straight, horizontal line:
\[\begin{align} f(x)&=0(x-h)^2+k \\ \\ f(x)&=k \end{align}\]
Considering the quadratic function \[f(x)=(x-3)^2+4,\] then its parabola will be opening upwards as the value of \(a\) is positive \((a=1)\).
Now, that you have learned about the effect of the value of \(a\) on the opening of a parabola, let's move to Translation.
Translation
Let's now have a look at the translation effects of constants \(h\) and \(k\) on parabolas.
First, let's introduce an important component of parabolas. The knowledge of this component will help you better monitor translations effected by those constants.
The vertex of a parabola is its lowest or highest point, located at \((h,k)\).
This point is also called turning point.
Here's a quick example to familiarize yourself with this term.
Considering the same function as above \[f(x)=(x-3)^2+4,\] now its vertex would be located at \((3,4)\).
We can now move to the effect of \(h\) and \(k\) on quadratic graphs, or parabolas.
Let's start by looking at how the value of \(h\), in the term \((x-h)\) of the standard form, represents a horizontal translation (or shift) of a parabola.
Let's examine figure 3 to see how this would work.
Fig. 3 - Horizontal translation effects of \(h\) on \(x^2\).
First, note that the standard form contains \((x-h)\), not \((x+h)\).
This means that \((x+5)\) will actually be a shift of the blue parabola 5 units to the left to the red parabola, as this equates to \(( x-(-5))\).
Similarly, \((x-5)\) will shift the blue parabola 5 units to the right to the green parabola.
Second, notice these translations did not affect the shape of the blue parabola; they just shifted all of its points to the left (green) or to the right (red).
So you can actually just focus on translating the vertex of the blue parabola, \((0,0)\), 5 units to the left, \((-5,0)\), or 5 units to the right, \((5,0)\), and then draw the parabola just like the initial one.
Now, similar to how \(h\) represents a horizontal translation, \(k\) represents a vertical translation.
This principle is exemplified in figure 4.
Fig. 4 - Vertical translations of \(k\) on \(x^2\).
Try some software such as desmos and experiment with quadratic graphs by changing their constants to grasp a better understanding of how they work.
Here, the sign of \(k\) tells you exactly where to move the parabola.
For \(k=5\), the blue parabola is shifted 5 units up to the green parabola, and for \(k=-5\), the blue parabola is shifted 5 units down to the red parabola.
Again, you can just focus on translating the vertex of the blue parabola, \((0,0)\), 5 units up, \((0,)\), or 5 units down, \((0,-5)\), and then draw the parabola just like the initial one.
Now that we have learned more about the effect of the value of \(h\) and \(k\) on the horizontal and vertical shift of the parabola from the origin, we are now ready to explore the last concept of Dilation.
Dilation
Let's now have a look at the dilation effect \(a\) can have on parabolas.
An absolute value of \(a\) greater than \(1\), \(|a|>1\), contracts a parabola, moving it closer to the \(y-\)axis;
An absolute value of \(a\) lower than \(1\), \(0<|a|<1\), dilates (or expands) a parabola, moving it away from the \(y-\)axis.
Once again, let's examine this transformation with figure 5.
Fig. 5 - The effects of coefficient \(a's\) value on contraction and dilation of \(x^2\).
For a value of \(a\) of \(10\) (absolute value greater than 1), the blue parabola is contracted to the green parabola, moving it closer to the \(y-\)axis. On the other hand, for a value of \(a\) of \(0.1\) (absolute value greater than 0 and lower than 1), the blue parabola is dilated to the red parabola, moving it away from the \(y-\)axis.
Having grasped the transformations that the constants \(a\), \(h\) and \(k\) have on quadratic graphs, you're now ready to plot a quadratic graph from scratch!
Graphing Quadratic Functions
We will summarize the graphing procedure using the steps below.
Step 1. Finding the coordinates of the vertex.
To go from an equation in standard form to a graph, you can begin by finding the coordinates of the vertex.
Has you've seen before, since translations move the whole graph from its starting position, you can use the vertex as a point of reference to monitor the shift of the whole graph. And this is because the shape of the graph remains the same.
Hence, for a function \[f(x)=a(x-h)^2+k,\] the \(x\)-coordinate of the vertex is \(h\) and the \(y\)-coordinate is \(k\).
Thus, the coordinates of the vertex are \((h,k)\).
Step 2. Determine the opening of the graph, and if there's dilation (or contraction).
The next step is to determine whether the graph opens upward or downward. This is determined by the \(a\) value:
- If \(a>0\), the graph opens upward;
- If \(a<0\), the graph opens downward.
Also, examine if there is dilation or contraction:
Step 3. Determine the \(x-\) and \(y-\)intercepts.
The final step before sketching the graph, is to determine the \(x-\) and \(y-\)intercepts.
- The \(x-\)intercepts are determined by setting \(y\), or \(f(x)\), to zero and solving for \(x\), i.e., \[y=0 \text{ or } f(x)=0.\] The solution to this equation (or solutions, if there are any) will be the \(x\)-coordinates of the \(x\)-intercepts;
- The \(y-\)intercept can be found by setting \(x\) to zero and solving for \(y\), i.e., \[f(0)=a(0-h)^2+k.\] The answer will be the \(y\)-coordinate of the \(y\)-intercept.
Finally, when sketching, ensure you go through all the graph's intercepts and the vertex. Label these coordinates on your graph.
Now, we move to the practical part of this explanation, let's consider an example!
Quadratic Graph Examples
How to sketch quadratic equations?
We first deal with an example showing us how to sketch graphs of quadratic equations.
Sketch the graph of the following quadratic,\[y=(x-2)^2-3.\]
Solution
Step 1. Determine the coordinates of the vertex.
We know the vertex is \((h,k)\) which is, therefore, \((2,-3)\).
Step 2. Determine the opening of the parabola, and if there's dilation.
We notice here that \(a=1\) and as this is positive the graph will be opening upwards. Also since \(a=1\) there is no contraction or dilation.
Step 3. Determine the intercepts.
We solve for our \(y-\) intercept by setting \(x=0\):\[\begin{align}y&=(x-2)^2-3\\ \\&=(-2)^2-3\\ \\&=4-3\\ \\&=1\end{align}\]Therefore our \(y-\) intercept is located at \((0,1)\).
Our \(x-\)intercepts are findable by setting \(y\) to 0. This leads to \(x-\)intercepts as follows:\[\begin{align}0&=(x-2)^2-3\\ \\ \pm \sqrt{3}+2&=x \\ \\ \therefore x&= 3.73 \mbox{ and } 0.27\end{align}\]This means our \(x-\)intercepts are located at \((0.27,0)\) and \((3.73,0)\). If we were to sketch all these points we would end up with the following graph.
Fig.6 - Graphic showing the sketch of the quadratic \(y=(x-2)^2-3\)
You may ask yourself now, do quadratic inequalities exist? and if the answer is yes, how do we graph them?
Well, quadratic inequalities are defined as all inequalities are defined, with the signs of inequality \(>, \ge, <, \le\).
In the next section, we shall learn more about them and their graphs.
Graphing Quadratic Inequalities
These have the same quadratic expressions as parabolas, except they have an inequality sign instead of an equal sign. This means that translations and contractions/dilations work the same as with the quadratic functions above.
There are some elements to learn which are unique to inequalities.
The first one is regarding the line of the graph, which can be dotted or continuous, depending on the direction of the sign of inequality:
Next, you must determine whether the shaded area (where the inequality is true) will be above or below the function. This, once again, depends on the direction of the inequality sign:
This information can be summarized in the table below.
| Area above the curve | Area below the curve |
Dotted line | \(>\) | \(<\) |
Continuous line | \(\ge\) | \(\le\) |
Although the method used in this explanation is where the inequality is true, you could shade the area on the other side so long as you specify (which you always should) what the shaded area represents.
The following graphs show some examples.
In figure 7 we see the area and lines for the inequalities \(y >x^2\) and \(y\ge x^2\).
Fig. 7 - Graphic showing conditions where left: \(y >x^2\) and right: \(y\ge x^2\).
Figure 8 shows the area and lines for the inequalities \(y<x^2\) and \(y\le x^2\).
Fig. 8 - Graphic showing conditions where left: \(y <x^2\) and right: \(y\le x^2\).
The images contain graphs where the shaded area represents where the equation holds true.
Although these graphs do not have any contractions or vertical and/or horizontal translations, these would work exactly the same as with any quadratic graphs.
Now, we move to an example of graphing quadratic inequalities.
How to sketch quadratic inequalities?
Now, we deal with an example showing us how to sketch graphs of quadratic inequalities.
Sketch the graph of the following inequality \[y<3(x+5)^2.\]
Solution
The first step is to write the quadratic equation,
\[y=3(x+5)^2.\]
Next, we proceed as we do for the quadratic graphs for equations. We can determine the vertex, the opening and the intercepts.
Determine the vertex.
The vertex for this quadratic will be \((-5,0)\).
Determine the opening.
Since \(a=3\) is positive, the graph will be opening upward.
Determine the intercepts.
Since there is no vertical translation for this function, its vertex will be its only intercept \(x-\)intercept.
Solving for the \(y-\)intercept by letting \(x=0\):\[\begin{align}y&=3(5)^2\\ \\&=75\end{align}\]
Hence the \(y-\)intercept is \((0,75).\)
If we let the shaded area be the area where the equation holds true, the shaded part will be below the curve. We can now graph the inequality with a dotted line, including all known points.
Fig. 9 - Graphic showing a sketch of the inequality \(y<3(x+5)^2\).
How to Determine the Function of a Quadratic Graph?
To go from a graph to an equation in standard form, you will need the vertex and any other point on the curve.
If the function is an inequality, the sign can be determined by the placement of the shaded area and the type of line.
Determine the equation of the following graph,
Fig. 10 - A graph to calculate the function from.
Solution
From the graph, the vertex is seen to be \((3,-5)\). Substituting these values into the standard form equation: \[f(x)=a(x-3)^2-5\]
To solve for \(a\), we can substitute the coordinates \((0,13)\):\[\begin{align}13&=a(0-3)^2-5\\ \\18&=9a\\ \\2&=a\end{align}\]
Hence, the equation for the graph is:\[f(x)=2(x-3)^2-5.\]
Quadratic Graphs - Key takeaways
- In the standard form, there are three constants that affect the way a parabola looks.
- By determining the intercepts and the vertex, we can switch between an equation and a graph of a quadratic.
- With the vertex and another point on a quadratic, we can solve the equation a graph has.
- The sign of an inequality determines where the shaded area is.
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