Lesson Video: Graphing Quadratic Functions | Nagwa Lesson Video: Graphing Quadratic Functions | Nagwa

# Lesson Video: Graphing Quadratic Functions Mathematics • Third Year of Preparatory School

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In this video, we will learn how to graph any quadratic function using a table of values and a given interval and identify the features of the graph.

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### Video Transcript

In this lesson, weβll learn how to graph any quadratic function using a table of values and a given interval and identify the features of the graph.

Now, quadratic equations are used in everyday life. Theyβre used in science, business, and engineering. For instance, they can determine the paths of moving objects, from bouncing balls to the flight path of a projectile. Businesses use them to forecast revenues and design packaging to minimize waste. We can also use quadratic equations to identify the minimum and maximum values of many different variables, such as speed, cost, and area. In particular, we can use the graphs of quadratic functions to easily determine information about maximum and minimum values and when the outputs will be equal to zero.

To do this, letβs begin by reminding ourselves what we mean when we say a function is quadratic. This is a single-variable polynomial function of degree two. In other words, a quadratic function has the form π of π₯ equals ππ₯ squared plus ππ₯ plus π for constants π, π, and π, where π cannot be equal to zero. Then, for this quadratic function π of π₯, π¦ equals π of π₯ is a quadratic equation.

Letβs now consider the graphs of two quadratic equations. Weβll begin by looking at π¦ equals π₯ squared plus four and a second quadratic equation π¦ equals negative π₯ squared plus nine. The graphs look a little something like this. And we might notice that they have a very similar shape. This shape is known as a parabola. In fact, all quadratics have this shape. And the sign of the constant value π, the coefficient of π₯ squared, tells us whether the parabola opens upwards or downwards. In particular, if the coefficient of π₯ squared is positive, if π is greater than zero, the parabola opens upward. And if π is less than zero, the parabola opens downward.

Letβs identify some of the other useful points on our graph, firstly the π¦-intercept, the point on each graph where it passes through the π¦-axis. In these cases, the π¦-intercepts are four and nine, respectively. But of course we might recall that we can find the value of the π¦-intercept by setting π₯ equal to zero and substituting that into each of our equations. But letβs look at what happens if we substitute π₯ equals zero into the general form of our quadratic function. We get π times zero squared plus π times zero plus π, which is simply equal to π. So, in fact, the value of the π¦-intercept of a quadratic function in this form is simply π. Or we can say it has coordinates zero, π.

In a similar way, we can in fact find the value of the π₯-intercepts if they exist by substituting π¦ equals zero or π of π₯ equal to zero and solving for π₯. Letβs demonstrate this in our first example.

Identify the π₯-intercepts of the quadratic function π of π₯ equals π₯ squared minus one.

And then we have a graph of that function shown. Now, in fact, there are two ways to answer this question then. We could simply look at the coordinates of the points where this graph passes through the π₯-axis. In fact, weβll use that technique to check our answer. Instead, we recall that the π₯-intercepts occur when π of π₯ or π¦ is set equal to zero. In this case then, we need to set the expression π₯ squared minus one equal to zero and solve for π₯. We could factor the expression on the left-hand side, or we can simply add one to both sides to get π₯ squared equals one. And then we can take the square root of both sides of this equation, remembering that in fact weβll need to take both the positive and negative square root of one. So π₯ is equal to positive or negative root one. But of course the square root of one is simply one. So π₯ is either equal to one or negative one.

We can denote these using subscript and work in ascending order. So π₯ sub one equals negative one and π₯ sub two equals one. Letβs check these on the graph. We begin by identifying the points at which the graph passes through the π₯-axis, as shown. Then, we look at the scale. In fact, we observe that four small squares is equivalent to two units. This means that two small squares must be equal to one unit. So these points are indeed negative one and one. The π₯-intercepts of the quadratic function given are π₯ sub one equals negative one and π₯ sub two equals one.

Now, before we move on, letβs inspect this graph a little more. We first notice that the turning point is in fact the minimum point. Itβs the point on the curve which outputs the smallest possible value of the function. We also see that if we construct a vertical line through this point, through the turning point of the graph, that line is the line of symmetry. And of course this point has a special name. Itβs called the vertex. And the vertex is really useful because it tells us the value of the minimum or maximum output of the function depending on the sign of the coefficient of π₯ squared. And we can note that since parabolas are symmetric in the vertical line through their axis of symmetry, this line must in fact lie halfway between the π₯-intercepts, if they exist. In fact, it will lie halfway between any π₯-values as shown on the graph with the same output.

Letβs summarize all of the properties of quadratic functions and their graphs. Remember, for the function π of π₯ equals ππ₯ squared plus ππ₯ plus π, if π is positive the graph opens upward and if itβs negative it opens downward. The π¦-intercept, which could be found by setting π₯ equal to zero, is actually just given by the point zero, π. And that can be zero, one, or two π₯-intercepts.

Now, these π₯-intercepts can be found if they exist by solving the equation π of π₯ equals zero. The single turning point of the function is called its vertex, and it will either be a maximum or a minimum. And of course these parabolas are symmetric in the vertical line through their vertex. We also saw that the axis of symmetry lies halfway between the π₯-intercepts, if they exist.

In our next example, weβre going to identify the axis of symmetry for a given quadratic curve.

Identify the axis of symmetry of the quadratic function π of π₯ equals π₯ squared minus one.

Remember, when we talk about the axis of symmetry of a quadratic curve, weβre talking about the vertical line that passes through its vertex or turning point. Now, whilst we could use the diagram to do so, weβre going to use an extra fact here. That is, the axis of symmetry will actually lie exactly halfway between the π₯-intercepts of the graph. In fact, it will lie halfway between any π₯-values which have the same output.

Letβs set π of π₯ equal to zero in order to find the value of the π₯-intercepts and then double-check this by looking at the graph. When we do, we get π₯ squared minus one equals zero. We add one to both sides and then take the positive and negative square root of one. So π₯ is positive or negative root one, and the two π₯-intercepts are negative one and one. And we can see they do indeed match the values on our graph.

We said that the axis of symmetry will actually lie exactly halfway between these two values. And of course we find their midpoint by finding their mean average. We add them together and divide by two. Thatβs negative one plus one, which is zero, divided by two, which is still zero. And so the axis of symmetry will pass through the point π₯ equals zero, and itβs a vertical line. Hence, the axis of symmetry has the equation π₯ equals zero.

Inspecting the graph, this does indeed look likely. But letβs check by choosing values of π₯ which have the same output. The point exactly halfway between negative four and four is zero, confirming to us that the axis of symmetry of our quadratic function is π₯ equals zero.

Now, up to this point, weβve used graphs and their equations to identify various features. In our next example, weβll demonstrate how we can actually use a table of values to help us sketch the graph of a quadratic function.

Which of the following graphs represents the equation π of π₯ equals negative two π₯ squared plus nine π₯ minus seven?

There are two ways we can answer this question. But the first thing that weβre going to do is establish whether the graph of our function opens upwards or downward. Firstly, we know that if we have the equation π of π₯ equals ππ₯ squared plus ππ₯ plus π and the value of π is positive, the graph opens upward. If itβs negative, it opens downward.

Now, for our function, the value of π is negative two. So itβs negative. This means the function opens downward. And we can therefore disregard any functions that open upward. Thatβs the function represented by graph (B) and the one represented by graph (E).

Now that weβve done this, weβre going to plot a table of values to identify the points through which this graph passes. Weβll need to do this for a variety of values of π₯. Letβs start with π₯ equals zero. Thatβs π of zero equals negative two times zero squared plus nine times zero minus seven. And thatβs negative seven. So when π₯ is equal to zero, the value of the function is negative seven. And that actually means that the π¦-intercept of our function is negative seven.

But of course the other technique we couldβve used is to identify that for the function ππ₯ squared plus ππ₯ plus π, the π¦-intercept has coordinates zero, π. Either technique here, substituting or identifying the feature of the graph, is equally acceptable.

Weβre now going to choose some positive and negative values of π₯ to satisfy each of the graphs weβve been given. First, letβs substitute π₯ equals negative three into the function. We get negative two times negative three squared plus nine times negative three minus seven. Thatβs negative 52. Similarly, letβs substitute π₯ equals negative two. When we do, we find that π of negative two is negative 33. And that means our graph must pass through the point negative two, negative 33. Substituting the remaining values, and we see that π of negative one is negative 18, π of one is zero, π of two is three, and π of three is two. Plotting the values that we can on our graphs, and we see we disregard option (C) completely. We notice itβs the wrong side of the π¦-axis. And in fact the graph that does pass through these points is graph (A).

Now, in fact, there was an alternative method here. We began by identifying the shape of the graph. We saw that it opened downward. And we found the value of its π¦-intercept. We could also have set π of π₯ equal to zero to find the value of the π₯-intercepts. Doing so wouldβve given us π₯ equals one and π₯ equals 3.5, which once again matches function (A). So the graph that represents the equation given is graph (A).

Letβs finish by recapping some of the key points from this lesson. In this lesson, we learned that a quadratic function is of the form ππ₯ squared plus ππ₯ plus π, where π is not equal to zero. We learnt that all quadratics are in the shape of a parabola, and they open upwards when π is positive and down when π is negative. And we learned that whilst we could set π₯ equal to zero to find the value of the π¦-intercept for the equation π¦ equals ππ₯ squared plus ππ₯ plus π, the π¦-intercept actually has coordinates zero, π.

We learned that the value of the π₯-intercepts, if they exist, can be found by solving π of π₯ equals zero. We learned also that graphs of quadratics have a single point called the vertex, and the sign tells us whether this will be a maxima or a minima. This vertex lies halfway between the π₯-intercepts. And of course quadratics are symmetric in the vertical line that passes through this vertex, which we call the axis of symmetry.

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