Claim. Every quadratic function has a U-shaped graph.
Proof. Consider the general quadratic function $ y = ax^{2} + bx + c $ .
Rewriting the equation in an equivalent form by completing the square, we obtain
\[ \begin{aligned}y&=ax^{2}+bx+c\\&=a\left(x^{2}+\frac{b}{a}x\right)+c\\&=a\left(x^{2}+\frac{b}{a}x+\frac{b^{2}}{4a^{2}}-\frac{b^{2}}{4a^{2}}\right)+c\\&=a\left[\left(x+\frac{b}{2a}\right)^{2}-\frac{b^{2}}{4a^{2}}\right]+c\\&=a\left(x+\frac{b}{2a}\right)^{2}+\left(c-\frac{b^{2}}{4a}\right).\\ \end{aligned} \]
This reveals that the general quadratic function is an algebraically transformed version of $ y = x^{2} $ . Consequently, its graph can be obtained from the U-shaped graph of $ y = x^{2} $ by a sequence of geometric transformations: a vertical stretch by a factor of a (with a vertical reflection if a < 0), then a horizontal shift, and finally, a vertical shift. Since a stretched U is still a U (albeit a skinnier or a fatter one), and a shifted U obviously is still a U, we may therefore conclude that the graph of $ y = ax^{2} + bx + c $ must always be U-shaped, as claimed.
One further observation: In the proof, we saw that if a, the quadratic’s leading coefficient, is negative, then the U will be reflected vertically. If this happens, it will, of course, open downwards. Otherwise (if the leading coefficient is positive) the U will open upwards. To sum up what we’ve discovered,
The graph of every quadratic function (i.e. function of the form $ y = ax^{2} + bx + c $ ) is U-shaped.
• The U opens upwards if the leading coefficient is positive, and downwards if it is negative.
• To graph a quadratic, we complete the square and apply the transformations revealed thereby.
Finally, a little terminology: The turning point in a U-shape is called its vertex. In the graph of a quadratic function, the vertex lies either where the function attains its maximum output value (if the graph opens downwards) or its minimum value (if it opens upwards).
Exercises
- Graph the following quadratics. Find the graphs’ vertices and their intersections with the axes.
\[ \begin{aligned}&a)y=x^{2}+8x+1\quad&b)y=2x^{2}+4x+2\quad&c)y=5x^{2}-3\quad&d)y=-x^{2}+10x-7\\&e)y=3x^{2}+4x+5\quad&f)y=-5x^{2}-12x+2\quad&g)y=\frac{3}{2}x^{2}+6x\quad&h)y=-\frac{3}{14}x^{2}+\frac{2}{7}x+1\end{aligned} \]
Can the graph of $ y = x^{2} $ be contained between two vertical lines? If so, which ones? If not, why not?
Many textbooks state that in the graph of $ y = ax^{2} + bx + c $ , the vertex’s x-coordinate will be -b/2a.
Explain why this is so.
Use this result to find the coordinates of the vertex of $ y = 5x^{2} + 4x - 1 $ .
Although this result will allow you to solve certain homework problems (like those in exercise 36) quickly, memorizing it is counterproductive. It will not help you learn mathematics. In contrast, each time that you graph a quadratic by completing the square, you reinforce two important mathematical techniques: completing the square and transformations. (Besides, if you want a shortcut, why not just use a computer?) The second reason for eschewing this formula is that once you’ve learned a little calculus, you’ll be able to find a quadratic’s vertex in seconds without having to rummage in your memory for anything.
If two rectangles have the same perimeter, must they have the same area? If so, explain why. If not, provide a counterexample.
Mr. Square plans to fence off a rectangular area in the middle of field. He has 100 feet of fencing. He dimly remembers learning that rectangles with the same perimeter can have different areas, and he wants his rectangle’s area as large as possible. Being Mr. Square, he’s pretty sure that the way to maximize the area is to make a square, but then, he’s been wrong before. Is he right this time? If not, why not? If so, prove it.
[Hint: Consider a rectangle with perimeter 100. Let x be the length of one of its sides. Express the rectangle’s area as a function of x. The function will be quadratic. Graph it, and think about its vertex in the context of Mr. Square’s problem.]
Mr. Square’s neighbor, Lana Evitneter, also plans to fence off a rectangular area with 100 feet of fencing. However, since a wall of her house will serve as one side of the rectangle, she actually needs fencing for just three sides. She asks Mr. Square to help her maximize the area of her rectangle. Naturally, Mr. Square recommends making a square. Is he right this time? If so, prove it. If not, find the dimensions of the rectangle that will actually maximize the enclosed area.
Mortified by his mistake, Mr. Square flees the neighborhood in his hot-air balloon. He takes off from the base of a very long hill, which has a constant slope of 1/3 (see the figure). The path that Mr. Square’s balloon takes happens to be the graph of $ y = -x^{2} + 4x $ . Alas, as you can see from the figure, his balloon doesn’t make it over the hill. Assuming that each unit on the axes represents 1000 feet,
What are the coordinates of the point at which the balloon lands?
How far (in feet) is the launch point from the landing point?
If the launch point is at sea level, then what is the balloon’s maximum altitude relative to sea level?
What is the balloon’s maximum altitude relative to the ground?
Parabolas
It has been observed that missiles and projectiles
describe a curved path of some sort. However, no
one has yet pointed out that this path is a parabola.
This… I have succeeded in proving.
- Galileo Galilei, Dialogues Concerning Two New Sciences, 3^{rd} Day, Introduction
The distance from a point to a line is, by definition, the length of the shortest path joining them: a straight path meeting the line at right angles. Thus, in the figure at right, the distance from point P to line AB is the length of segment PQ.
Draw a line and a point on a piece of paper. Call the point F and the line d. Locate a point equidistant from F and d, and mark it on the page. Then, find and mark as many other points as you can that are equidistant from F and d. After a while, you’ll have a picture that looks something like the one at right. In your mind’s eye, picture the curve that passes through each and every one of the infinitely many points equidistant from F and d. This curve is called a parabola.
Definition. A parabola is the set of all points equidistant from some fixed point (called the parabola’s focus) and some fixed line (called the parabola’s directrix).
While all parabolas are U-shapes, very few U-shapes are parabolas. If you draw a random U-shape on a page, it will almost certainly not be a parabola. Try it: First draw a random U-shape on a page, then try to guess where its focus and directrix would be if it were a parabola. Now start checking points on the U (preferably with the aid of a ruler). If you can find even one point on the U that is not exactly the same distance from your prospective focus and directrix, then your U-shape is not a parabola – at least not with those choices of focus and directrix. Difficult as it is to draw a reasonably accurate circle freehand (without the aid of a compass), it is still more difficult to draw a reasonably accurate parabola.
Mathematicians have studied parabolas for well over two thousand years on account of their remarkable geometric properties. Amazingly, in the early 17 $ ^{th} $ century, Galileo proved that parabolas have physical significance as well: A projectile moving under the influence of gravity alone will always follow a parabolic path. When you toss a ball to your dog, the path that the ball follows when it leaves your hand is not merely U-shaped, but parabolic: As it moves through the air, the ball remains equidistant from an invisible focus and directrix. Take a physics class, and you’ll learn why.
We can discover a parabola’s equation by translating its definition into algebraic terms. This will be especially easy if we place the axes so that the parabola’s vertex is at the origin and its focus lies on the positive y-axis. This setup ensures that the focus’s coordinates will be $ (0, p) $ for some positive number p. Moreover, since the vertex is p units from the focus, it must also (by the parabola’s definition) lie p units up from the directrix; hence, the equation of the directrix must be y = -p.
Problem. Derive the equation of the parabola with vertex $ (0,0) $ and focus $ (0, p) $ .
Solution. We seek an equation satisfied by the coordinates of all the parabola’s points. Let $ (x, y) $ be a variable point on the parabola. Its distance to the focus $ (0, p) $ is, by the distance formula,
\[ \sqrt{(\Delta x)^{2}+(\Delta y)^{2}}=\sqrt{\boldsymbol{x}^{2}+(\boldsymbol{y}-\boldsymbol{p})^{2}}, \]
while its distance to the directrix is $ y + p $ . (It lies y units above the x-axis, which itself lies p units above the directrix, as in the figure above.)
The two distances we’ve just computed are, by the parabola’s definition, equal. That is,
\[ \sqrt{x^{2}+(y-p)^{2}}=y+p. \]
This equation is satisfied by our variable point (and thus by every point) on the parabola, so it is the parabola’s equation. To polish it, we square both sides of the equation and then simplify. The resulting polished equation of the parabola is, as you should verify, $ y = ( ) x^{2} $ .
The equation of the parabola with vertex $ (0,0) $ and focus $ (0,p) $ is
\[ \boldsymbol{y}=\frac{1}{4\boldsymbol{p}}\boldsymbol{x}^{2} \]
We’ll use this fact to establish two little preliminary results that we’ll then parlay into a big theorem. Here’s the first little result.
Claim 1. The graph of $ y = x^{2} $ is not merely U-shaped, but parabolic.
Proof. The equation $ y = x^{2} $ has the form in the box above, with p = 1/4. Hence, its graph is a parabola with vertex $ (0, 0) $ and focus $ (0, 1/4) $ . (And its directrix is y = -1/4.)
The second of our two preliminary results concerns stretched parabolas. Suppose we stretch a parabola. The result will certainly be U-shaped, but will it still be parabolic? Remarkably, it turns out that if we stretch any parabola by any factor in any direction (not just vertically or horizontally), the result will still be a parabola! Although this full statement is too difficult for us to prove here, we can easily prove one very special – and very useful – case.
Claim 2. If we stretch the graph of $ y = x^{2} $ vertically, the result is still a parabola.
Proof. If we stretch the graph of $ y = x^{2} $ vertically by a factor of k, its new equation will be $ y = kx^{2} $ . This matches our boxed equation for a parabola with $ k = 1/(4p) $ ; or equivalently, with p = 1/4k. Hence, the graph of $ y = kx^{2} $ is indeed a parabola with vertex $ (0, 0) $ and focus $ (0, 1/(4k)) $ .
With our two preliminary results squared away, let’s turn to our big theorem.
Theorem. The graphs of all quadratic functions are parabolas.
Proof. In an earlier section (“Graphs of Quadratic Functions”), we proved that any quadratic’s graph can be obtained from the graph of $ y = x^{2} $ by following a certain sequence of transformations: first a vertical stretch (sometimes accompanied by a reflection), and then some shifts.
Since then, you’ve learned that the graph of $ y = x^{2} $ is not just a U, but a parabola (Claim 1), and that a vertical stretch will preserve its parabolic nature (Claim 2). What about reflections and shifts? Might they disrupt a stretched parabola’s parabolic nature, turning it into a mere U-shape? No, this won’t happen. After all, reflections and shifts don’t change a graph’s shape at all (they only change the graph’s location), so they must obviously preserve a parabola’s parabolic nature, too.
To sum up, we’ve now seen that the graph of each and every quadratic function can be obtained from one specific parabola $ (y = x^{2}) $ by some sequence of “parabola-preserving” transformations. It follows that the graph of every quadratic function must be a parabola, as claimed.
Combining this last theorem with our earlier work on graphing quadratics, we may conclude that
The graph of every quadratic function (i.e. function of the form $ y = ax^{2} + bx + c $ ) is a parabola.
• The parabola opens up if the leading coefficient is positive, and down if it is negative.
• To graph a quadratic, we complete the square and apply the transformations revealed thereby.
Exercises
Pick a random number. Call it n. Square it. The point $ (n, n^{2}) $ lies on the graph of $ y = x^{2} $ . Compute its distances to the parabola’s focus and directrix, and verify that these are indeed equal.
Is $ y = 2x^{2} $ the graph of a parabola? If so, find its vertex, focus, and directrix.
Is $ y = ()x^{2} $ the graph of a parabola? If so, find its vertex, focus, and directrix.
Find the equation of the parabola with vertex $ (0,0) $ and focus $ (0,5) $ .
Find the equation of the parabola with vertex $ (0,0) $ and focus $ (0,1/5) $ .
Prove that if we stretch the graph of $ y = x^{2} $ horizontally, the result will still be a parabola.
Find the focus and directrix of the parabola whose equation is $ y = 3x^{2} - 12x + 13 $ .
[Hint: Complete the square, think about how each geometric transformation affects the focus & directrix.]
- Find the focus and directrix of the parabola whose equation is $ y = -3x^{2} - 6x + 1 $ .
The Reflection Property of Parabolas
At any point P on a parabola, draw the parabola’s tangent there. Next, draw the line segment joining P to the focus. Finally, draw the ray from P that is parallel to the parabola’s axis of symmetry. We can prove that the line segment and ray make equal angles with the tangent. This so-called “reflection property” of parabolas does not hold for other U-shapes: only parabolas. Esoteric though it may seem, this parabolic reflection property has remarkable physical consequences, which you’ll be able to appreciate after reading the following paragraphs about optics.
Light rays bounce off of flat surfaces in a simple manner: When a ray strikes a flat surface, it reflects off at the same angle at which it struck. The figure at right depicts a light ray striking (and then departing) a reflective surface at an angle of $ 24^{} $ .
If a light ray strikes a curved surface, the same rule holds, but now the angles are measured between the light ray and the tangent to the surface where the ray hits it. You can see why this is so if you imagine zooming in on the point of tangency so closely that the curve and the tangent line become indistinguishable. This virtual identity of a curve and its tangent line – when viewed at a microscopic scale – is, incidentally, a major theme of calculus.
Now let us return to the reflection property of parabolas. If light leaves a parabola’s focus and hits the parabola at point P, in what direction will it be reflected? If you’ve understood the three preceding paragraphs, you should be able to convince yourself of the following: Because of the reflection property, the ray that strikes the parabola at P will “bounce off” the parabola parallel to the parabola’s axis of symmetry. This being so, suppose that we illuminate a bulb at the parabola’s focus, so that light streams out of it in all directions. Amazingly, all the light that hits the parabola will bounce back in the same direction: parallel to the parabola’s axis of symmetry! It is for this reason that parabolic mirrors are used in flashlights, headlights, and so forth.
Parabolic mirrors can also be used in reverse to concentrate parallel rays into a point. For example, solar rays that reach us on Earth are effectively parallel, since the Earth and Sun are so tiny compared to the vast distance between them. If we capture solar rays in a parabolic mirror, we can concentrate them into one very hot point at the mirror’s focus. (Hence the name focus, which means “hearth” in Latin.) You can find images of such “burning mirrors” or “solar furnaces” online, including the world’s largest, located in a French village in the Pyrenees. Another example: Satellite receiving dishes are parabolic in shape to concentrate the satellite’s signals into the dish’s focus, where the dish’s transmitter is located.
Proof of the Parabola’s Reflection Property
Our proof will make use of the following fairly obvious geometric fact.
The perpendicular bisector of a line segment AB divides the plane into three sets of points: those on A’s side of the bisector (which are closer to A than B); those on B’s side of the bisector (which are closer to B than A); and those on the bisector itself (which are equidistant from A and B);
For example, in the figure at right, we must have PA = PB, but QB < QA.
Claim. Light emitted from a parabola’s focus reflects off the parabola parallel to its axis of symmetry.
Proof. First we’ll set the stage. Let P be any point on a parabola.
Now draw segment FP joining it to the focus, and ray $ PD’ $ parallel to the axis of symmetry. Among all the straight lines through P that do not enter angle $ FD’ $ , its obvious that only one makes equal angles with FP and $ PD’ $ . I’ve drawn this line $ (PS’) $ in gray. Next, we extend ray $ PD’ $ backwards until it hits the directrix at D. Since $ PD’ $ is parallel to the axis of symmetry, the extension must be perpendicular to the directrix. Finally, let $ FD’ $ intersection with the gray line be called S. The stage is now set.
In the proof’s first act (of three), we will prove that the gray line is $ FD^{} $ ‘s perpendicular bisector. The key is to show that $ FPS DPS $ , which we can do as follows: First, FP = PD by the parabola’s defining property. Next, $ FS = DS $ (since both angles are equal to $ D’S’ $ ; one by the gray line’s definition, the other by vertical angles). Finally, SP is common to both triangles. Therefore, by SAS-congruence, $ FPS DPS $ as claimed. It follows that FS = DS. That is, the gray line bisects FD. Moreover, $ FP = DP $ , and since they form a straight line, these equal angles must be right angles. Thus the gray line is indeed $ FD’ $ ’s perpendicular bisector.
In act two, we’ll prove the gray line is tangent to the parabola by showing that all the parabola’s points (besides P, of course) lie on one side of the gray line. To this end, let Q be any other point on the parabola, and drop perpendicular QR to the directrix. By the parabola’s definition, QF = QR, which is less than QD (in any right triangle, leg < hypotenuse), so QF < QD. Thus, Q is on F’s side of FD’s perpendicular bisector, the gray line. Since Q was an arbitrary point, all the parabola’s points (besides P) lie on F’s side of the gray line, which is thus tangent to the parabola at P, as claimed.
The third and final act: Imagine that a ray of light emitted from the parabola’s focus F strikes the parabola at P, as in the figure above. As explained earlier, the angle at which the ray is reflected equals the angle at which it strikes the parabola’s tangent at P. But this tangent is the gray line. Since the ray strikes this line at angle $ FS $ , it reflects off at an angle of the same magnitude. But by definition of the gray line, that angle is $ D’S’ $ . That is, the light reflects off the parabola along ray $ PD’ $ , which, by definition, is parallel to the parabola’s axis of symmetry.
