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Graphs of quadratic polynomials

Graphs of quadratic polynomials
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We’re now going to explore some aspects of the geometry of quadratic polynomials. We’ll see that once you understand this geometry, some common questions are rather easy to answer. So we have a general quadratic polynomial, ax squared plus bx plus c. We’ll suppose that its leading coefficient, the a parameter, is strictly positive. What do you think its graph looks like? Well, recall that we had found a way to rewrite our quadratic using this notation with the quantity and braces that you see and where the delta means the discriminant.
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Now once you’ve rewritten it this way, it jumps out at you that it’s going to be the same graph as the graph of x squared, except the graph is going to be squished or stretched. It’s going to be shifted up or down or left or right because you see the x has become 2ax plus b. That is, there’s been a dilation and a translation of the argument, and then the values have also been translated and dilated. So the graph is going to look like our parabola graph for x squared. And it’s going to open up because the a is strictly positive. If the a were negative, it would be a downward opening parabola. And now we see some cases arise.
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If the discriminant is strictly positive, there are two roots of our quadratic. That means the graph must intersect the horizontal axis in two places, like so. If delta is strictly negative, there are no roots, so, therefore the graph of the function must lie above the x-axis everywhere. And finally, in the case when delta is 0, there’s exactly one root, so this must be a case where we have just exact contact between the parabola and the x-axis at one point. Let’s see how this appreciation of the geometry can lead to a very easy answer to a question such as this, which might seem hard otherwise.
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For what values of x is it true that x squared minus 4x plus 2 is negative? Well, we recognise that we have a quadratic polynomial on the left. We calculate its discriminant. Delta turns out to be 8, and, therefore, it’s strictly positive, so there are two roots. We can find them by the quadratic formula, and here they are. Now when there are two distinct roots and a is positive, what is the graph of the function going to look like? Answer, it’s going to intercept the x-axis in two points. What are those two points, those values of x where we have these intercepts? Answer, of course, they’re the two roots of the quadratic, which we’ve calculated.
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And then we can answer the question just visually. For what values of x is it true that this function is negative, that is, the graph is below the x-axis? Well it’s obviously the closed interval determined by the two roots, and that’s the answer. Here’s another example of a type of question we can easily answer if we bear in mind the geometry of quadratic polynomials. I give you one, ax squared plus bx plus c. The a is strictly positive. And I ask, what is the value of x for which this function attains its minimum, its lowest value? Well, we have this expression that we so carefully worked out before, alternate expression for a quadratic polynomial.
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We know the graph is going to look like this, and we’re looking for the value of x that corresponds to the lowest point in its graph. Where will that be achieved? Well, if you look at the quantity in braces, you can see that it’s something positive, and then there’s a constant. So it will be least when the something positive is 0, that is, when 2ax plus b is equal 0. That’s when x is equal minus b over 2a. So this special value of x, which I’ve labelled x m, means the value of x for which you’ll be at the lowest point of the parabola. Let’s take an example. You want to minimize the function, x squared minus 4x plus 1.
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You want to find the minimum value of that function, and you want to identify the point where this minimum is attained. We calculate x m. It turns out, in this instance, to be 2. Now the function has value minus 3 when x is 2, as you can check. And so the minimum value of the function is minus 3. Now here’s an example of a somewhat more subtle nature. We again want to calculate the minimum of a quadratic polynomial, however, we now restrict the study to the interval 1, 2. How do we proceed? Well, we calculate the minimizing point for our quadratic function, turns out minus b over 2a, in this case, to be 1/2.
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And we notice that 1/2 is not in the interval over which we wish to take the minimum. That is, 1/2 is not in the interval 1, 2. If it were in the interval, then, of course, that would be the minimizing point, but it isn’t. In fact, it’s to the left of the interval as you can see. The fact that it’s to the left of the interval allows us, geometrically, to see that the function, f, is increasing on the interval 1, 2. It gets bigger as you move to the right. Therefore, if you’re interested in minimizing f, you want to be at the leftmost point of the interval 1, 2, that is, at x equals 1.
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That must give you the minimum relative to that interval. So we calculate f at the point 1 and the minimum is 2. We are going to continue fun with algebra in the next segment when we do polynomial division.
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Precalculus: the Mathematics of Numbers, Functions and Equations

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