Graph each function. See Examples 1 and 2. ƒ(x) = -½ x²

Question

Accepted Answer

Welcome back. I am so glad you're here. We are asked to plot the given function. Our given function is G of X which equals negative 1/ X squared. Then we have a blink graph. We have a vertical Y axis, a horizontal X axis. They come together at the origin. The domain for what's drawn for our X axis is from negative five to positive five. And the range for what's drawn for our Y axis is from negative 0.5 to positive 0.1. All right. So the first thing we want to do is take a look at our given function and determine its base function. So what's happening to the X here? And here we see that our X is being squared. So our base function is Y equals X squared. We recall from previous lessons that that is an upward opening parabola with its vertex at the origin. So it's got one sort of arm heading up toward positive infinity for the Y values but negative infinity for the X values. And this isn't going to be to scale and then it curves down and intercepts at the origin where its vertexes and then turns around and heads back up curves to positive infinity for the X values and the Y values in quadrant one. So there's our Y equals X squared. What's different about our given function? Well, two things. One, the X squared is being multiplied by 1/12 and it's also being multiplied by a negative, we look at those separately. So what do they mean? Well, the 1/12 is not being multiplied directly to that X, it's happening after the X has been squared. So when it's not happening directly to the X, before that base function takes place, it indicates a vertical change. And in this case, because 1/12 is between zero and one, it indicates vertical shrinking. Now, the other change here is this negative being multiplied again, that's not happening directly to the X. It happens after the X has already been squared. So in this case, that indicates reflection over the X axis. So those are our two changes. The vertical shrinking will really be able to see just because of the way this graph is scaled with the Y axis and the reflection over the X axis, we just take that Parabola and we flip it so that instead of heading to positive infinity for the Y values, it's heading toward negative infinity for the Y values. So we've got a downward opening Parabola with one arm of it heading toward negative infinity for X and Y still has its vertex at the origin. And then the other arm heads toward negative infinity for Y but positive infinity for X. So that's a general idea of what we've got going on. But we can make it more precise, we can create an XY table looking at our graph. We see what the domain is for X. So we choose some X values, plug those into our function negative 1 12 multiplied by X squared discern the Y values and plot those points. So for the X values here, I'm going to choose negative 20 and positive two. Now plugging in negative two for X, we have negative 1 12 multiplied by not X squared but negative two squared, negative two squared is a positive four. So we have negative 1 12 multiplied by four which is a negative one third erasing this other graph. Now we can plot these points for the point negative two, negative one third from the origin. We'll go to the left two and then we'll go down one third which is going to be just past negative 0.3. All right. Now for the X value of zero, we've negative 1 12 multiplied not by X squared but by zero squared. Well, zero squared is zero. So we have negative 1 12 multiplied by zero which is zero. There's our point at the origin. All right, last point positive two for X we have negative 1 12 multiplied by two squared. Two squared is four. So negative 1 12 multiplied by four. Again, that is negative one third. So for the 0.2 negative one third from the origin, we go to the rate two and then go down one third. Now we can connect these points for our downward facing parabola. We're heading toward negative infinity for X and Y and passing through that point of negative two, negative one third and then it's going to curve at the origin, turn around and then head through that other point. A positive two negative one third. Mine's not super smooth then heading toward negative infinity for the Y values and positive infinity for the X values. And there is our G of X equals negative 12 X squared. Well done. We'll catch you on the next one.

Graph each function. See Examples 1 and 2. ƒ(x) = -½ x²

Key Concepts

Quadratic Functions and Their Graphs

Effect of the Leading Coefficient

Plotting Key Points and Vertex

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