Question Video: Identifying Graphs of Quadratic Equations | Nagwa Question Video: Identifying Graphs of Quadratic Equations | Nagwa

Question Video: Identifying Graphs of Quadratic Equations Mathematics • Second Year of Secondary School

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The graphs of three quadratic equations are shown on the diagram below. Find the equation for each graph.

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Video Transcript

The graphs of three quadratic equations are shown on the diagram below. Find the equation for each graph.

In this question, we’re given the graph of three quadratic equations. And we need to determine the equation for each individual graph. And in fact, we know a lot of different ways of determining the equation of a quadratic function from its graph. So the first step we need to determine is which of these methods should we use? We’ll do this by looking at the diagram. The first thing we can notice from the diagram is that we can determine the coordinates of the vertices of each of the three graphs from the diagram. And if we can determine the coordinates of the vertex of a quadratic function, then we can use this to determine its equation in vertex form. And this is one method which would definitely work and give us the correct answer.

However, we’re going to use a second method by noticing all three of the graphs pass through the same two 𝑥-intercepts. We can then recall if a real number 𝑟 is a root of a quadratic function, then 𝑥 minus 𝑟 is a factor of that quadratic function. And remember, a root of a quadratic function will be its 𝑥-intercept when it’s graphed. Therefore, if we have two distinct roots of our quadratic function 𝑟 sub one and 𝑟 sub two, we can write this in the form 𝑎 times 𝑥 minus 𝑟 sub one multiplied by 𝑥 minus 𝑟 sub two. This is sometimes called the factored form.

And in our case, all three of the graphs have the same two roots, one and negative one. So we can substitute 𝑟 sub one is equal to one and 𝑟 sub two is equal to negative one into this expression. And this means all three of the functions of these graphs will be of the form 𝑎 times 𝑥 minus one multiplied by 𝑥 plus one. And it’s worth pointing out the question does ask us for the equation of these graphs, so we’ll need to set these equal to 𝑦.

Let’s start by determining the value of 𝑎 in the blue diagram. The first thing we can note is 𝑥 squared minus one factors by using a difference between squares to give us 𝑥 minus one times 𝑥 plus one. So we can rewrite the equation of the blue quadratic function as 𝑦 is equal to 𝑎 times 𝑥 squared minus one. We can then distribute 𝑎 over the parentheses to get 𝑎𝑥 squared minus 𝑎.

We want to determine the value of 𝑎. And the easiest way to do this is the substitute 𝑥 is equal to zero into the equation. And to do this, we need to notice that when 𝑥 is equal to zero, we’ll output the 𝑦-intercept of the curve. For the blue curve, this is negative one. In other words, when 𝑥 is zero, 𝑦 will be equal to negative one. This gives us negative one is equal to negative 𝑎. And we can multiply through by negative one to see that 𝑎 is equal to one. Therefore, we’ve shown the equation of the quadratic function in blue is 𝑦 is equal to 𝑥 squared minus one.

So let’s clear some space and follow the same process to determine the equation of the quadratic curve given in red. Once again, its equation is in the form 𝑦 is equal to 𝑎 times 𝑥 minus one multiplied by 𝑥 plus one because it has roots at one and negative one. And by distributing our parentheses or by recalling the difference between squares formula, we can rewrite this as 𝑦 is equal to 𝑎𝑥 squared minus 𝑎.

And now, we can do exactly the same thing we did before. We want to substitute 𝑥 is equal to zero into this equation. When we do this, the output will be the 𝑦-intercept of the curve. In this case, the 𝑦-intercept is negative four. This then gives us that negative four is equal to negative 𝑎. And if we multiply this equation through by negative one, we get that 𝑎 is equal to four. Therefore, the equation of the quadratic function given in red is 𝑦 is equal to four 𝑎 squared minus four.

Let’s now clear some space and determine the equation of the final curve. In exactly the same way, the green curve will have the equation 𝑦 is equal to 𝑎𝑥 squared minus 𝑎 for some real value of 𝑎. We want to substitute 𝑥 is equal to zero into this equation. This will output the 𝑦-intercept of the curve, which we can see is negative nine. Therefore, negative nine is equal to negative 𝑎. And we can multiply through by negative one to see that 𝑎 is equal to nine, which we can substitute back into our equation to see the equation of the green curve is 𝑦 is equal to nine 𝑥 squared minus nine. And this is our final answer.

Therefore, we were able to show the equation of the blue curve is 𝑦 is equal to 𝑥 squared minus one. The equation of the red curve is 𝑦 is equal to four 𝑥 squared minus four. And the equation of the green curve is 𝑦 is equal to nine 𝑥 squared minus nine.

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