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ACT Math Intermediate Algebra Drill 1, Problem 3. Can you find the value of a in this expression?
Algebra and Functions: Drill 4, problem 2. If the function for the area of a circle with respect to its diameter was graphed in the first quad...
Algebra and Functions: Drill 4, Problem 3. The sum of the quadratic parent function and the line given has which of the following as its graph?
SAT Math 9.5 Algebra and Functions 213 Views
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Description:
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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