Question Video: Identifying the Equation of a Graph After a Reflection Mathematics

Find 𝑔(π‘₯), where the graph of the function 𝑔(π‘₯) is a reflection across the π‘₯-axis of the graph of the function 𝑓(π‘₯) = 2π‘₯ + 5.

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Video Transcript

Find 𝑔 of π‘₯, where the graph of the function 𝑔 of π‘₯ is a reflection across the π‘₯-axis of the graph of the function 𝑓 of π‘₯ equals two π‘₯ plus five.

Let’s begin by thinking about what’s happening here. We’re taking a function 𝑓 of π‘₯, and we’re reflecting it across the π‘₯-axis, and then we get that function 𝑔 of π‘₯. Our job is to find the equation for the function 𝑔 of π‘₯. So, let’s remind ourselves how we achieve a transformation of a reflection across the π‘₯-axis. Given the graph of the function 𝑦 equals 𝑓 of π‘₯, 𝑦 equals negative 𝑓 of π‘₯ is a reflection of that original graph across the π‘₯-axis. In other words, to achieve the reflection, we take the original function and we multiply the entire expression by negative one. In this case then, 𝑔 of π‘₯ must be negative 𝑓 of π‘₯, which is negative two π‘₯ plus five. Let’s multiply each term in the expression two π‘₯ plus five by negative one, and that gives us negative two π‘₯ minus five.

So, we’ve worked out that the graph of 𝑔 of π‘₯ is the equation 𝑦 equals negative two π‘₯ minus five. We can check this by sketching both graphs on the coordinate plane. The graph of 𝑦 equals two π‘₯ plus five will look a little something like this. It passes through the 𝑦-axis at five. That’s its 𝑦-intercept, and it has a slope of two. This means for every one square right we move, we need to move two squares up to hit the line. This means it also passes through the point one, seven. Then the graph of 𝑦 equals negative two π‘₯ minus five will look like this. This time its 𝑦-intercept is negative five, and it has a slope of negative two. So, for one square right, we move two squares down, taking us to a point with coordinates one, negative seven.

We can indeed see that these are reflections of one another across the π‘₯-axis or the line 𝑦 equals zero. So, 𝑔 of π‘₯, which is a reflection across the π‘₯-axis of the graph of 𝑓 of π‘₯, has equation negative two π‘₯ minus five.

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