Okay. So let's give this problem a try. Here we have the function f of x, which is equal to c times x squared, and we're asked to graph our function f of x when c is equal to 2 and when c is equal to 1 half. Now, what we have on the graph already here is a curve that represents our function f of x is equal to x squared. This is the general case for a parabola, but we need to figure out what's going to happen if we take various constants and plug them into this function. So, let's begin. For the case where our constant here is equal to 2, this means that fx=2∙x2. This is what the function is going to look like if we plug 2 into the constant.
Now, I'm going to try a bunch of different x values. First, I'll try an x value of 0. If this happens, we would have 2 times 0 squared, and 2 times 0 is 0; 0 squared is just 0. So, this means we're going to have a point at 0. Next, I'll plot a value of -1. If I do this, we're going to get 2 times negative one squared, just replacing this x with a -1. In this case, 2 times -1 is -2, so we'll have -2 squared, and -2 squared is actually 4. Thus, at a value of -1, we're going to be at a y-value of 4. Lastly, I'll try a point of 1. So, in this case, we'll have 2∙x2, where x = 1, giving a value of 4. So our graph is going to look something like this when we replace this constant with a 2. This makes sense because we discussed previously that having a constant multiplied inside of your function causes a horizontal shrink if the constant is greater than 1.
But now let's try the situation where our constant is 1 half. This means our constant is between 0 and 1, so let's see what happens. In this case, our function f(x) is going to become 12∙x2. Well, let's begin by plugging in an x value of 0. In this case, we have 1 half times 0 squared, and anything multiplied by 0 is just 0, so we have the same origin point at 0. Now, I'll try an x value of 2. In this case, we get 1 half times 2 squared. The 1 half and 2 will cancel each other out, giving you 1 squared, which is just 1. So at an x value of 2, you would be at a y value of 1, and similarly, for -2, you would again be at a y value of 1. This results in a horizontal stretch because our constant is between 0 and 1, which differs from the horizontal shrink we had previously.
It's interesting to note that when we had a horizontal shrink, it appeared as if there was a vertical stretch. Likewise, when we had a horizontal stretch, it seemed as if there was a vertical compression or shrink. These appearances are often the case in symmetric graphs like parabolas, suggesting a correlation between horizontal and vertical transformations.
So, this is how you solve the problem and understand the effects of constants on the graph of a quadratic function. Hopefully, you found this explanation helpful.