Question Video: Finding the Equation of a Function After a Reflection Mathematics

The parabolic graph represents a function 𝑔(π‘₯) after reflection on the π‘₯-axis. Find the original function 𝑓(π‘₯).


Video Transcript

The following parabolic graph represents a function 𝑔 of π‘₯ after reflection on the π‘₯-axis. Find the original function 𝑓 of π‘₯.

In order to answer this question, let’s begin by finding an equation that describes the function 𝑔 of π‘₯. Now, it’s a parabolic graph. So, what does that tell us about its equation? Well, it’s going to be a quadratic equation, and that’s of the form 𝑦 equals π‘Žπ‘₯ squared plus 𝑏π‘₯ plus 𝑐. To have this sort of inverted parabola, this n shape, we know that the value of π‘Ž is going to be negative. It’s less than zero. We also know that the value of 𝑐 gives us the 𝑦-intercept of the graph. Here, the value of 𝑐 must therefore be equal to negative six. So, our equation is going to be of the form 𝑦 equals π‘Žπ‘₯ squared plus 𝑏π‘₯ minus six.

Now there’s a couple of different ways we can find the values of π‘Ž and 𝑏. One is to choose two more coordinates that lie on our curve and form and solve a system of linear equations. The other is to consider the vertex. We’re going to do the former method, and then we’ll use the vertex to check our answer. Since four small squares represents two units, then we know that two small squares represents one unit. So, our curve passes through the points one, negative five and two, negative six. In other words, when π‘₯ is one, 𝑦 is negative five. And when π‘₯ is two, 𝑦 is negative six.

So, our first equation using the coordinates one, negative five is negative five equals π‘Ž times one squared plus 𝑏 times one minus six. This simplifies to negative five equals π‘Ž plus 𝑏 minus six. Then adding six to both sides, we get one equals π‘Ž plus 𝑏. And this is our first equation. Let’s repeat this process with the next pair of coordinates. We get negative six equals π‘Ž times two squared plus 𝑏 times two minus six, which simplifies to negative six equals four π‘Ž plus two 𝑏 minus six. We can add six to both sides of this equation, and then we can simplify by dividing through by two. So, we have our second equation. It’s zero equals two π‘Ž plus 𝑏.

Let’s clear some space and solve the system of linear equations. Since the coefficient of 𝑏 in each equation is the same, we’re going to subtract equation one from equation two. When we do, two π‘Ž minus π‘Ž is π‘Ž, 𝑏 minus 𝑏 is zero, and zero minus one is negative one. So, we find that π‘Ž equals negative one. Let’s now substitute this value of π‘Ž into our first equation. So, negative one plus 𝑏 equals one. And adding one to both sides, we find that 𝑏 is equal to two. We’re now ready to substitute this back into our equation, giving us 𝑦 equals negative π‘₯ squared plus two π‘₯ minus six. And remember, we’re expecting a negative value for π‘Ž since our graph is inverted.

Now we said that we can check this by considering the vertex of the graph: the point one, negative five. If we write our answer in completed square form, this gives us indication as to the value of the vertex. So, let’s do that just to double-check. We begin by factoring negative one out of the first two terms. Then, when we complete the square on the expression π‘₯ squared minus two π‘₯, we get π‘₯ minus one all squared, and then we subtract one squared, which is one. Redistributing negative one and we get negative π‘₯ minus one all squared plus one minus six. And so, the equation, alternatively, can be written as 𝑦 equals negative π‘₯ minus one all squared minus five.

These two values here, these constants, give us indication of the vertex of our graph. We change the sign of the term inside the parentheses to give us the π‘₯-coordinate. And we see that the turning point has coordinates one, negative five as we expected. So, we now know the equation of the graph given. So, what happens when we reflect this graph over the π‘₯-axis? Well, suppose we have some function 𝑦 equals 𝑓 of π‘₯. 𝑦 equals negative 𝑓 of π‘₯ is a reflection of the original graph of the function across the π‘₯-axis. Now, of course, we’ve got a function 𝑔 of π‘₯ that’s already been reflected across the π‘₯-axis. It follows that we can reverse this process by reflecting again, so the process is going to be the same.

Redefining 𝑔 of π‘₯ as negative π‘₯ minus one all squared minus five, we can say that 𝑓 of π‘₯ is going to be equal to negative 𝑔 of π‘₯. Essentially, we need to take our function 𝑔 of π‘₯ and multiply it by negative one. So it’s negative one times negative π‘₯ minus one all squared minus five. And if we redistribute those parentheses, multiplying by negative one, we get π‘₯ minus one squared plus five. We’ll leave this in completed square form, and we find that 𝑓 of π‘₯ is π‘₯ minus one all squared plus five.

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