Six Basic Graphs

You should learn the shapes of the following graphs so well that they – like circles and lines – appear before your mind’s eye whenever you encounter their equations:

Thinking about each of these preceding equations for a few minutes should suffice to explain their graphs’ shapes. Do this before you move on. Then do the same for the next three graphs, whose shapes you should also learn:

Exercises

  1. Does the graph of $ y = $ ever cross the line y = 10? If so, at what point does it cross?

  2. Why does the graph of $ y = $ have points in the third quadrant, unlike the graph of $ y = $ ?

  3. To attain large y-values for either of the functions in the last problem, one needs to put in very large x-values.

In other words, $ y = $ and $ y = $ both grow very slowly.

Which of these two slowpokes grows faster than the other? Give some numerical evidence for your claim.

  1. Among $ y = x^{2} $ , $ y = x^{3} $ , and $ y = |x| $ , which grows fastest? Which grows slowest?

  2. Does the graph of y = 1/x ever touch either axis? How can you be certain?

  3. There is one point that lies on all six of the graphs above. What are its coordinates?

  4. Sketch the basic shapes of the graphs of $ y = x^{4} $ , $ y = x^{5} $ , $ y = x^{6} $ , and $ y = x^{7} $ .

  5. Describe the relationship between the graphs of y = x, y = -x, and $ y = |x| $ .

I ntersections

The intersections of two graphs are the points lying on both of them. A simple method to find them is based directly on the Fundamental Principle of Coordinate Geometry.

Example 1. Where does the line y = 2x intersect the ellipse $ x^{2} + 9y^{2} - 9 = 0 $ ?

Solution. Let $ (a, b) $ represent any point that lies on both graphs.

Since it lies on the $ ellipse’ $ ‘s graph, its coordinates must, by the FPCG, satisfy the $ ellipse’ $ ’s equation. Thus, we know that

\[ a^{2}+9b^{2}-9=0. \]

Similarly, since it lies on the line’s graph, the FPCG tells us that

\[ b=2a. \]

A-ha! We now have two equations in two unknowns, a situation we discussed in the last chapter. Recall the process: First, we’ll combine the two equations into a single equation in one unknown. From there, we can just follow our noses to the problem’s end. In this case, we can combine the equations very easily, by substituting the second equation into the first, which yields

\[ a^{2}+9(2a)^{2}-9=0. \]

Solving this, we get $ a = / $ . These are the intersections’ x-coordinates. Then, since these points lie on the line, we can find their y-coordinates by plugging each x-coordinate into the line’s equation, $ y = 2x $ . When we do that, we find that the y-coordinates are $ / $ .

Thus, the line intersects the ellipse at points (3/√37, 6/√37) and (−3/√37, −6/√37).

Such is the process for finding intersections. Its logic, I hope, is crystal clear. That said, we can speed the process up: From a strictly computational standpoint, introducing $ (a,b) $ into the problem is superfluous, and we can cut to the chase by combining the graphs’ equations without substituting a and b for x and y, proceeding to the solution from there. $ ^{†} $ Introducing a and b as an intersection point’s coordinates has psychological value, since it helps us understand the logic behind the technique, but in hindsight, once we’ve digested that logic, we see that it isn’t computationally necessary. We may therefore abandon those training wheels and proceed to the faster, simplified procedure.

We can summarize the process concisely if we use some new terminology: A system of two equations is a pair of equations for which we seek “simultaneous solutions” – i.e. points that satisfy both equations. To solve a system of two equations means to find all such points. $ ^{†} $

Finding the Intersections of Two Graphs.

To find the intersections of two graphs, we solve the system of their equations.

Solving a system of two equations in two unknowns is a three-step process. First, combine the equations via some sort of substitution so as to produce an equation in one unknown. Second, solve this equation. Third, substitute its solution back into one of the original equations and obtain the remaining unknown.

Example 2. Find the intersection points of the circle and line at right.

Solution. Because the circle is centered at the origin and its radius is 2, its equation is $ x^{2} + y^{2} = 4 $ . Since the line has slope 2 and y-intercept 0, its equation is y = 2x. We want to find the points whose coordinates satisfy both equations.

To find them, we’ll solve the system of these two equations.

First, we’ll combine the two equations into a single equation involving just one unknown. We can do this by substituting the second equation into the first, yielding $ x^{2} + (2x)^{2} = 4 $ .

Second, we solve this equation. Doing this (I’ll leave the details to you) yields $ x = / $ . These are the intersection points’ x-coordinates.

Finally, to obtain the corresponding y-coordinates, we substitute these x-coordinates back into the line’s equation, whereby we find that $ y = / $ .

We therefore conclude that the line and circle intersect at $ (, ) $ and $ (, ) $ .

Example 3. Find the intersection points of the two circles at right.

Solution. Clearly, the larger circle’s equation is $ x{2}+(y-2){2}=4 $ , while the smaller circle’s is $ x{2}+y{2}=1 $ .

First, we must combine these two equations in two unknowns into one equation in one unknown. We can accomplish this by rewriting the small circle’s equation as $ x^{2} = 1 - y^{2} $ , and substituting this into the large circle’s equation to get $ (1 - y^{2}) + (y - 2)^{2} = 4 $ .

Second, we solve this equation. Doing so (you should supply the details) reveals that y = 1/4. This tells us that the y-coordinate of both intersection points is 1/4.

Finally, we obtain the points’ x-coordinates by substituting y = 1/4 into the small circle’s equation, thus obtaining an equation in x alone: $ x^{2} + (1/4)^{2} = 1 $ . Solving it yields $ x = /4 $ .

Thus, the intersection points of the two circles are $ (/4, 1/4) $ and $ (-/4, 1/4) $ .

When solving intersection problems, you may spell out the details as I did in example 1, or make use of the accelerated procedure we used in examples 2 and 3. The choice is yours; either way, be sure you understand why that method works.

Exercises

  1. Sketch the graphs of the following equations, and find the point(s) where they intersect:

\[ a)y=3x-2,\quad y=-2x+3 \]

\[ \mathsf{b})5x+3y=3,\quad2x-7y=1 \]

\[ \mathsf{c})x^{2}+y^{2}=1,\ y=-\frac{2}{3}x \]

\[ \mathsf{d})\left(x-2\right)^{2}+\left(y+1\right)^{2}=3,\quad x+3y=3\qquad\mathsf{e})y=x^{3},\quad y=3x \]

\[ \mathsf{f})y=x^{2},\ y=2x+1 \]

\[ \mathfrak{g})x^{2}+y^{2}=1,\ y=x^{2} \]

  1. $ y=,y=x/2 $

\[ i)y=\sqrt{x},\quad x=4000000 \]

  1. $ y = $ , $ y = x + 1 $ [Hint: Review Exercise 39.]
  1. Mr. Anonymous tries to find the intersections of the lines $ y = 2x + 1 $ and $ y = 2x + 2 $ as follows:

First, he equates the two expressions for y, obtaining $ 2x + 1 = 2x + 2 $ .

He then subtracts 2x from both sides, obtaining the absurd statement 1 = 2.

What happened? What does this mean? Did Mr. Anonymous make a mistake somewhere? Discuss.

  1. Later, Mr. Anonymous attempts to find the intersections of $ x^{2} + y^{2} = 1 $ and $ y = x + 2 $ .

He begins by substituting the linear equation into the circle’s equation.

After some algebraic manipulation, he ends up with the quadratic equation $ 2x^{2} + 4x + 3 = 0 $ . He finds that this quadratic has no real solutions, as you should verify.

What does this lack of solutions mean geometrically?

  1. The figure at right shows the graphs of two lines with their corresponding equations.

  1. Do they cross? If so, where? If not, why not?

  2. Find the equation of the line passing through $ (0,0) $ and perpendicular to the upper line in the figure.

[Hint: Recall Exercise 22.]

  1. Where does the line whose equation you found in part (b) cross the lower line in the figure?
  1. The hyperbola in the figure at right is the graph of y = 1/x.

Find the coordinates of the four points at which it intersects the circle.

[Hint: You will end up with a $ 4^{th} $ -degree polynomial equation that you must solve. Don’t give up. Look closely and you will see that it is really a quadratic in disguise.]

  1. A line segment’s perpendicular bisector is the unique line that is perpendicular to the segment and passes through its midpoint. Around 300 B.C., Euclid proved that all three perpendicular bisectors of any triangle’s sides intersect at a single point. This point, moreover, turns out to be triangle’s circumcenter, the center of the unique circle (the triangle’s “circumcircle”) that passes through all three of the triangle’s vertices.

Your problem: Find the equation of the unique circle passing through the points…

  1. $ (0, 0) $ , $ (3, 0) $ , $ (0, 2) $

  2. $ (0, 0) $ , $ (4, 0) $ , $ (-2, 2) $

  3. $ (0, 0) $ , $ (2, 4) $ , $ (6, 2) $ .

I ntersections (encore)

One common technique for finding intersections is particularly well-suited to the equations of lines. The trick is simple and is built upon just two ideas.

We know, of course, that multiplying an equation’s sides by the same nonzero constant produces an equivalent equation – that is, an equation with the same solutions. But if two equations are satisfied by the same coordinate pairs, then they have the same graph. This observation yields our first big idea:

Multiplying the sides of an equation by the same nonzero constant preserves its graph.

Suppose, for example, that we’re working with the graph of $ 2x + 5y = 1 $ . If we realize that replacing this equation with $ 6x + 15y = 3 $ would be more convenient (for some algebraic reason or other), we can safely make this algebraic change without affecting the underlying geometry. $ ^{*} $

Next, consider a balanced scale, with objects of equal weight already resting on its two pans. If we add an additional pound of gold on one pan, and 16 ounces of gold on the other, the scale will obviously remain balanced, because adding equals to equals preserves the equality. Or, restated algebraically, if a = b and c = d, then $ a + c = b + d $ . (The same idea holds for subtraction.) This is our second big idea:

Adding (or subtracting) the sides of two valid equations produces a third valid equation.

Combining these two ideas leads to our new intersection-finding technique: Multiply each equation’s sides by well-chosen constants, then add or subtract the resulting equations. The key is to choose the constants so that the subsequent addition or subtraction eliminates one of the two variables. Let’s see an example.

Example. Find the point $ (x, y) $ where the lines $ 5x + 3y = 3 $ and 2x - 7y = 1 intersect.

Solution. First, multiply both sides of the first equations by 2 and both sides of the second by -5. The lines’ graphs are preserved, but their new equations, $ 10x + 6y = 6 $ and $ -10x + 35y = -5 $ , suit our purposes better, because adding their corresponding sides will eliminate x. Thanks to those equal but opposite $ 10x $ terms, adding yields an equation with just one variable: 41y = 1.

From this, we can read off the intersection’s y-coordinate: 1/41. Feeding this into either of the original equations yields its x-coordinate: 24/41. Thus, the lines intersect at $ (24/41, 1/41) $ .

Exercises

  1. In the preceding example, we “multiplied the equations” by 2 and -5, but other values would work, too. Think of another promising pair of numbers; use them to re-solve the problem, and verify that the solution is the same.

  2. Sometimes, it suffices to multiply just one equation’s sides before adding or subtracting. For example, try solving the system $ 2x + y = 3 $ and $ 5x + 3y = 3 $ by multiplying the first equation’s sides by 3, then subtracting.

  3. Use the new technique described above to find the intersections of the following pairs of lines.

  1. $ 3x + 5y = 2 $

$ 6x - 4y = 1 $

\[ \begin{aligned}&2x-7y=5\\&3x+7y=4\end{aligned} \]

\[ \begin{aligned}c)19x-3y&=4\quad&d)x+y&=1\\-4x-6y&=1\quad&2x-y&=1\end{aligned} \]