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SAT Math 9.5 Algebra and Functions 213 Views
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Description:
SAT Math 9.5 Algebra and Functions
SAT Math 9.5 Algebra and Functions
Transcript
- 00:02
Here's a question that's part of a balanced Shmoopy breakfast.
- 00:06
If f (x) = x squared – 6x + 9, and f (x + 1) = 1, what is one possible value of x?
- 00:16
Well, it wants us to find one possible value of x, which just means… solve for x.
- 00:21
The requirements for a value of x are: f of x is equal to x squared - 6x + 9, and f of x + 1 = 1.
- 00:31
We’ll need to put together the two requirements to find x.
Full Transcript
- 00:37
Let’s take a closer look at the second requirement.
- 00:40
This just means that if we plugged the value (x + 1) into the function,
- 00:44
which we’re given in the first requirement, we can solve for x.
- 00:47
To solve this equation, we’re first going to have to turn it back into a polynomial.
- 00:51
So, we expand (x + 1) squared first.
- 00:55
By applying foil, we turn (x + 1) squared into x squared + 2x + 1.
- 01:03
Then, we distribute -6 into the second parentheses. We get -6x - 6.
- 01:10
Now we can combine like terms. There’s only one x squared term, so that stays by itself.
- 01:15
However, we have both 2x and -6x, so we can combine those to -4x.
- 01:20
Then, all of the constants add to +3. Great, we have a polynomial!
- 01:24
To be more specific… it's a quadratic.
- 01:27
We can just stick this puppy into the quadratic formula and come out with the answer.
- 01:31
Plugging in our values, we get that x is equal to 4 plus or minus the square root of 16 – 12 over 2.
- 01:40
This simplifies to 4 plus or minus 2 over 2. The possibilities are 1 or 3.
- 01:47
That's our answer either 1 or 3.
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