+2446 In order to figure this one out, we will utilize some bits of manipulation.
The y-intercept of a polynomial function is determined by the constant term of each polynomial. We know, however, that the y-intercept is located at (0,−5). Knowing this, we have already eliminated the last answer choice, y=x4+4x3−2x2−5x because its y-intercept is located at the origin.
The next rule I will use is something called the Descartes' rule of signs. We know, based on the given info, that at least one zero of this polynomial is negative.
For the three remaining candidates, let's plug in the function f(−x) and count the number of sign changes. If you don't mind, I converted all the equations to function notation.
| Candidate Function 1 | Candidate Function 2 | Candidate Function 3 | |
| Function | f(x)=x3−4x2+2x−5 | f(x)=x3+4x2−2x−5 | f(x)=x3+4x2+2x−5 |
| Calculate f(−x) | f(−x)=−x3−4x2−2x−5 | f(−x)=−x3+4x2+2x−5 | f(x)=−x3+4x2−2x−5 |
| Number of Sign Changes in f(−x) | 0 | 2 | 2 |
Why am I doing this, you may ask? Well, the Descartes' rule of signs states that the number of sign changes in f(−x) equals the amount of negative zeros a function has or less than by an even number.
We can make a table to show to show the amount of negative zeros that each function can have. I'll put it in table format for your convenience.
| f(x)=x3−4x2+2x−5 | f(x)=x3+4x2−2x−5 | f(x)=x3+4x2+2x−5 | |
| Number of Negative Zeros | 0 | 2 or 0 | 2 or 0 |
Wow, we have eliminated y=x3−4x2+2x−5 from the candidates because it has no real negative roots. However, this is even more helpful than it may appear at first. Now, we will finally plug in our first point, (−1,0) and see if results in a true statement. I'll choose x3+4x2+2x−5.
| y=x3+4x2+2x−5 | Plug in y=0 and x=-1. These are friendly numbers, so this should be a quick process. |
| 0=(−1)3+4(−1)2+2∗−1−5 | Evaluate this as true or false. |
| 0=−1+4−2−5 | Continue simplifying. |
| 0=−4 | This is an untrue statement. |
This is very useful information. I now know that (−1,0) does not satisfy y=x3+4x2+2x−5, which means that this point cannot lie on this line. By process of elimination, this means that y=x3+4x2−2x−5, the second answer choice from the top, is the correct equation.
+9488 Maybe it is supposed to be
f(x) = 2(2)x - 4 ???
If so...
f(0) = 2(2)0 - 4 = 2(1) - 4 = 2 - 4 = -2 So the graph passes through (0, -2) .
f(1) = 2(2)1 - 4 = 2(2) - 4 = 4 - 4 = 0 So the graph passes through (1, 0) .
The first graph passes through (0, -2) and (1, 0) , so it looks right.
Here's a graph of y = 2(2)x - 4
