Lesson Video: Quadratic Functions in Different Forms Mathematics

In this video, we will learn how to evaluate and write a quadratic function in different forms.

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Video Transcript

In this video, we will learn how to evaluate and write a quadratic function in different forms.

We recall, firstly, that a quadratic expression is one in which the highest power of the variable that appears is two. So, the expression includes a squared term and no terms with higher powers, such as cubed or to the power of four. For example, the expression π‘₯ squared plus two π‘₯ minus three is a quadratic expression, as is two 𝑦 squared minus three. The expression π‘Ž minus one all squared is also a quadratic expression. Because once we’ve distributed the parentheses, we see that this is equivalent to π‘Ž squared minus two π‘Ž plus one, and so the expression includes an π‘Ž squared term.

There are a number of different key forms in which a quadratic function can be written. The first of these is the general or expanded form. 𝑓 of π‘₯ equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐, where π‘Ž, 𝑏, and 𝑐 are constants and π‘Ž must not be equal to zero. It’s usual to write the terms in order of descending powers of the variable. So, we have the π‘₯ squared term first, then the π‘₯ term if there is one, and finally the constant term if there is one. Although this isn’t essential.

The second form is completed square or vertex form. 𝑓 of π‘₯ equals π‘Ž multiplied by π‘₯ plus 𝑝 all squared plus π‘ž, where π‘Ž, 𝑝, and π‘ž are all constants and, again, π‘Ž must be nonzero. The form we use depends on which features of the quadratic function or its graph we’re particularly interested in. In this video, we’ll mostly be focusing on the completed square form because it’s useful for determining the coordinates of the vertex and the equation of the axis of symmetry of a quadratic graph.

The final form in which we can express a quadratic function is its factored or factorized form. 𝑓 of π‘₯ equals π‘Žπ‘₯ plus 𝑏 multiplied by 𝑐π‘₯ plus 𝑑, where π‘Ž, 𝑏, 𝑐, and 𝑑 are constants. Although, it isn’t possible to express every quadratic in this form. Let’s now consider some examples. In our first example, we’ll see how we can use the completed square or vertex form of a quadratic function to determine the coordinates of the vertex of its graph.

Find the vertex of the graph of 𝑦 equals five multiplied by π‘₯ plus one all squared plus six.

Here, we have a quadratic function. And it’s been expressed in its completed square or vertex form. In general, this form is π‘Ž multiplied by π‘₯ plus 𝑝 all squared plus π‘ž. So, comparing the general form with our quadratic, we see that the value of π‘Ž is five, the value of 𝑝 is one, and the value of π‘ž is six. We were asked to use this form to find the vertex or turning point of the graph of this quadratic function.

Now, we should recall that all quadratic functions have the same general shape, which is called a parabola. When the value of π‘Ž is positive, the parabola will curve upwards. And when the value of π‘Ž is negative, the parabola will curve downwards. The vertex of the graph is its minimum point in the first instance and its maximum point in the second. Our value of π‘Ž in this question is five, which is positive. So, we’re in this first instance here. We’re looking for the coordinates of the minimum point of this graph. Let’s consider then how we can minimize this quadratic function and for what value of π‘₯ this minimum will occur. And then, we’ll see how we can generalize.

Let’s consider each part of the function in turn, starting with π‘₯ plus one squared. Well, this is a square. And we know that squares are always nonnegative. Whatever value of π‘₯ we choose, once we have added one and then squared it, the result will be zero or above. So, the minimum value of π‘₯ plus one all squared is zero. We then multiply this by five. But, of course, anything multiplied by zero is still zero. So, the minimum value of five multiplied by π‘₯ plus one all squared is still zero. We then add six, so the entire function increases by six units, which means that its minimum value overall is six. This gives the minimum value of the function or the minimum value of 𝑦. So, it is the 𝑦-coordinate of the vertex.

We then need to determine what value of π‘₯ causes this to be the case. Well, it’s the value of π‘₯ such that π‘₯ plus one all squared takes its minimum value of zero. It’s, therefore, the value of π‘₯ such that the expression inside the parentheses, that’s π‘₯ plus one, is equal to zero. The solution to this simple linear equation is π‘₯ equals negative one. So, we find that the π‘₯-coordinate of the vertex is negative one. The coordinates of the vertex of this graph then are negative one, six.

Now, let’s consider how we can generalize. The π‘₯-coordinate of the vertex was the value that made the expression inside the parentheses zero. In the case of the general expression π‘₯ plus 𝑝, this will be the value negative 𝑝 because negative 𝑝 plus 𝑝 give zero. So, this is the π‘₯-coordinate of vertex. The 𝑦-coordinate of six was the value that was added on to this squared expression. So, in the general form, that will be the value π‘ž. So, we can generalize. If a quadratic function is in its completed square or vertex form 𝑦 equals π‘Ž multiplied by π‘₯ plus 𝑝 all squared plus π‘ž, then the coordinates of its vertex are negative 𝑝, π‘ž.

Now, the quadratic function in this question was given to us in its completed square form. In our next example, we’ll remind ourselves how to take a quadratic in its general form and write it in its completed square form.

In completing the square for the quadratic function 𝑓 of π‘₯ equals π‘₯ squared plus 14π‘₯ plus 46, you arrive at the expression π‘₯ minus 𝑏 all squared plus 𝑐. What is the value of 𝑏?

Here, we’re given a quadratic in two forms, its general or expanded form and its completed square form. These two expressions describe the same quadratic, and therefore they’re equal. We can, therefore, form an equation π‘₯ squared plus 14π‘₯ plus 46 is equal to π‘₯ minus 𝑏 all squared plus 𝑐.

Now, there are two approaches that we can take to answering this question. The two approaches involve working in opposite directions. In our first approach, we’ll manipulate the expression on the right-hand side of our equation to bring it into its expanded form. Using whatever method we’re most comfortable for squaring a binomial, we see that π‘₯ minus 𝑏 all squared is equivalent to π‘₯ squared minus two 𝑏π‘₯ plus 𝑏 squared. So, the expression on the right-hand side becomes π‘₯ squared minus two 𝑏π‘₯ plus 𝑏 squared plus 𝑐.

As these expressions are equivalent for all values of π‘₯, we can now compare coefficients on the two sides of the equation. The coefficients of π‘₯ squared on each side are one. And then, if we compare the coefficients of π‘₯, we have 14π‘₯ on the left-hand side and negative two 𝑏π‘₯ on the right-hand side, giving the equation 14 equals negative two 𝑏. We can solve this equation for 𝑏 by dividing each side by negative two, giving 𝑏 equals negative seven.

Now, we have actually completed the problem because all we we’re asked for was the value of 𝑏. But suppose we’d also been asked to determine the value of 𝑐. We could do this by comparing the constant terms on the two sides of the equation. On the left, we have positive 46. And on the right, we have 𝑏 squared plus 𝑐. So, that gives us a second equation; 𝑏 squared plus 𝑐 is equal to 46. We know that 𝑏 is equal to negative seven and negative seven squared is 49. So, substituting this value into our equation, we have 49 plus 𝑐 equals 46. And subtracting 49 from each side, we find that 𝑐 is equal to negative three.

This means that the completed square or vertex form of our quadratic, using 𝑏 equals negative seven and 𝑐 equals negative three, is π‘₯ minus negative seven all squared minus three. Of course, π‘₯ minus negative seven is better written as π‘₯ plus seven. So, we can express this as π‘₯ plus seven all squared minus three.

So, that’s our first method in which we expanded to the expression on the right-hand side. In our second method, we’ll see how we can work the other way. So, we’ll start on the left-hand side and bring it into its completed square form. We already know what we’re working towards. It’s π‘₯ plus seven all squared minus three. Now, notice that the value inside the parentheses of positive seven is exactly half the coefficient of π‘₯ in our original equation. And this is no coincidence. This will always be the case. So, we begin by writing π‘₯ squared plus 14π‘₯ plus 46 as π‘₯ plus seven all squared. We’ve halved the coefficient of π‘₯ to give the value inside the parentheses.

The trouble is, though, π‘₯ plus seven all squared isn’t just equivalent to π‘₯ squared plus 14π‘₯; it’s equivalent to π‘₯ squared plus 14π‘₯ plus 49. So, we’ve introduced an extra 49. We, therefore, need to subtract this so that the expression we have is still equivalent to π‘₯ squared plus 14π‘₯. This new expression of π‘₯ plus seven all squared minus 49 is, therefore, exactly equivalent to π‘₯ squared plus 14π‘₯. We also need to include the positive 46 so that the two sides of the equation are the same.

Now, that value 49 is, of course, the square of seven. So, what we’re doing is subtracting the square of the value inside our parentheses. The final step is just to simplify. We have negative 49 plus 46, which is equivalent to negative three. And so, we’ve found that this quadratic, in its completed square form, is π‘₯ plus seven all squared minus three, which is the same as we found using our first method.

Comparing this then to the given form of π‘₯ minus 𝑏 all squared plus 𝑐, we would see that positive seven is equal to negative 𝑏. Dividing or, indeed, multiplying both sides of this equation by negative one, we find that 𝑏 is equal to negative seven. Using two methods then, that’s working in both directions, we’ve found that the value of 𝑏 is negative seven.

Completing the square is a little bit more complicated when the value of π‘Ž, that’s the coefficient of π‘₯ squared in the general form, is not equal to one. So, let’s consider an example of this.

Which of the following is the vertex form of the function 𝑓 of π‘₯ equals two π‘₯ squared plus 12π‘₯ plus 11. And we’re given five answer options.

So, we have a quadratic function in its general or expanded form, and we’re asked to find its vertex form. That’s 𝑓 of π‘₯ equals π‘Ž multiplied by π‘₯ plus 𝑝 all squared plus π‘ž, where π‘Ž, 𝑝, and π‘ž are constants we need to find. To answer to this question then, we need to write our quadratic function in its completed square form.

Now, this is slightly trickier than usual because the coefficient of π‘₯ squared in our function isn’t one. It is two. Now, we can deal with this by factoring by this coefficient. You can either factor from all three terms giving two multiplied by π‘₯ squared plus six π‘₯ plus 11 over two. Or we can simply factor this coefficient from the first two terms, giving two multiplied by π‘₯ squared plus six π‘₯ plus 11. And personally, I think the second method is easier.

What we’re now going to do is complete the square on the expression within the parentheses. That’s π‘₯ squared plus six π‘₯. We begin by halving the coefficient of π‘₯ to give the number inside the parentheses. So, we have π‘₯ plus three all squared. And then, we need to subtract the square of this value. So, we have π‘₯ plus three all squared minus nine. A quick check of redistributing these parentheses confirms that π‘₯ plus three all squared minus nine is indeed equal to π‘₯ squared plus six π‘₯.

Now, we need to be very careful here. All of this expression of π‘₯ squared plus six π‘₯ was being multiplied by two. So, we need to put a large set of brackets or parentheses around π‘₯ plus three all squared minus nine, multiply it by two. And then, we still have the 11 that we were adding on. So, we found that our function 𝑓 of π‘₯ is equivalent to two multiplied by π‘₯ plus three all squared minus nine plus 11.

Next, we need to distribute the two. So, we have two multiplied by π‘₯ plus three all squared and then two multiplied by negative nine, which is negative 18, and then plus 11. Remember that 11 is not being multiplied by two. Finally, we just simplify negative 18 plus 11 is negative seven. So, we have our quadratic in its vertex form two multiplied by π‘₯ plus three all squared minus seven.

Looking carefully at the five answer options we were given because they’re all quite similar, we see that this is answer option (a). Now, in this question, we just worked through the completing-the-square process ourselves and then determined the correct answer option. It would actually have been possible to eliminate a couple of the options straightaway though. If we look at the vertex form of the function, we see that within the parentheses we have just π‘₯ plus 𝑝 all squared. We could, therefore, have eliminated options (b) and (c) as they are not the vertex form of any function because inside the parentheses they each have two π‘₯.

We could also have eliminated option (e) because we can see that there is no two involved in this option. And as the coefficient of π‘₯ squared in our original function was two, we’d need a factor of two outside the parentheses as we have in answer options (a) and (d). The only difference between options (a) and (d) is the sign inside the parentheses. We have positive three for option (a) and negative three for option (d). We should remember, though, that the sign here is always the same as the sign of the coefficient of π‘₯ in the original function. So, in this case, it’s positive. In any case, we have our answer though. The vertex form of this function is 𝑓 of π‘₯ equals two multiplied by π‘₯ plus three all squared minus seven.

Quadratic functions can be used to model practical situations such as the path followed by an object or a worded problem. In our final example, we’ll see how we can form a quadratic function from a description.

Two siblings are three years apart in age. Write an equation for 𝑃, the product of their ages, in terms of π‘Ž, the age of the youngest sibling.

So, we’re told in this question to use the letter π‘Ž to represent the age of the younger sibling. We now want to find an expression in terms of π‘Ž for the age of the older sibling. Well, we’re told that the siblings are three years apart in age. So, the older sibling is three years older than the younger. An expression for the age of the older sibling is, therefore, π‘Ž plus three. The product of their ages means we need to multiply them together. So, we take our two expressions of π‘Ž and π‘Ž plus three and multiply them.

This is a quadratic equation. Because if we were to distribute the parentheses, we have π‘Ž multiplied by π‘Ž giving π‘Ž squared and π‘Ž multiplied by three giving three π‘Ž. Either of these two forms would be acceptable for the answer, but we’ll give our answer as 𝑃 equals π‘Ž multiplied by π‘Ž plus three.

Let’s now review some of the key points from this video. Firstly, one of the key forms in which a quadratic can be written is its general or expanded form. 𝑓 of π‘₯ equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐, where π‘Ž, 𝑏, and 𝑐 are constants and π‘Ž must be nonzero. We can also express quadratics in their completed square or vertex form. 𝑓 of π‘₯ equals π‘Ž multiplied by π‘₯ plus 𝑝 all squared plus π‘ž, where π‘Ž, 𝑝, and π‘ž are constants and, again, π‘Ž must be nonzero.

We’ve also seen that for a quadratic function expressed in its completed square or vertex form, then its vertex, which will be a minimum when the value of π‘Ž is positive and a maximum when the value of π‘Ž is negative, is at the point with coordinates negative 𝑝, π‘ž. Through our examples, we saw how we can convert between these two key forms either by following the completing-the-square process or by distributing the parentheses and simplifying the resulting expression. Although we didn’t need to use it in this video, we also know that some quadratic functions can be written in a factored or factorized form.

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