Graphs, Functions, and Transformations
This topic seems impossibly complicated, but you can get through it if you do one simple thing: ignore all the fancy terms and just act like a robot.
Here is an example of a function: . For every value of x you put in the right side of the equation, the function spits out one (and only one) corresponding value. Pretty robotic. Let’s jam x = 0 into the function and see what comes out.
So if x is zero, the corresponding value of f(x) is 5.
Now, on a graph, f(x) acts as the y-value, so you’ve just discovered the point (0, 5). Seem familiar? It should, since that’s point B, your y-intercept on page . The function is nothing more than the equation for line AC in the form y = mx + b, only f(x) is masquerading as y.
Think about it. If you understand linear equations in the form y = mx + b, then you understand how a linear function appears on a graph. Granted, not every function is going to be a straight line: you can get some funky functions if you start squaring terms. In general, though, whatever function you’re given, put in values for x and see what values for y are spit out. Then graph them.
Transformations can be tricky. Again, acting robotic can make them easier. You’ll be given an initial function, and then this function will be “transformed” in some manner. Here’s an example using a linear function:
7. Which line shows the function f(x) = 3x +1 after f(x) has been transformed to f(x – 1)?
(A) line m
(B) line n
(C) line o
(D) line p
(E) line q
All right, robot, here are your instructions:
  1. Take (x –1) and place it into the original function.
  2. Solve this new equation.
  3. Using the new slope and y-intercept of this “transformed function,” locate the new line.
You can replace f(x) with y anytime you want.
The y-intercept (b in y = mx + b) is –2, so the new, transformed line has a point at (0, –2). Only line p crosses that point, so D is the answer.
That’s a transformation. Good work. Power down, robot.
Imagine if the world’s strongest person took a line, grabbed it in both hands, and then somehow bent it. The resulting shape would be a parabola, a U-shaped curve that can open either upward or downward on a coordinate graph.
Parabolas occur when the x value in a function is squared. A common equation for a parabola is: . The equation for our parabola is y = 4x2 – 6x +3.
You can see this is similar to the equation for a line, , only we’ve added a squared term to the front of it.
Parabolas have many characteristics. The two key ideas you have to learn for the SAT are how to:
  1. Determine whether the parabola opens upward or downward
  2. Locate the vertex—the bottom point of the U—of the parabola
To determine whether the parabola opens upward or downward, look at the a value (the number in front of the squared term). The parabola shown in the diagram conforms to the equation . The a value, 4, is positive, which is why the parabola is heading upward.
The vertex is a bit trickier. This point is found by the very complicated formula:
It would be nice if this formula was at the front of every Math section, but no such luck. For the parabola , the vertex ends up at . If you don’t believe us, block out about five minutes of time and work it out yourself.
Thankfully, that’s the last geometry concept we’re going to cover. Now that your mind is crammed with the right facts, it’s time to show you the best way to put this knowledge to use.
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