First, factor out a 3.
3(x² - 9)
In any quadratic ax² + bx + c, we can split the bx term up into two new terms which we want to equal the product of a and c.
In this case, we have x² + 0x - 9. (the 0x is a placeholder)
We want two numbers that add to 0 and multiply to get -9.
Obviously, these numbers are 3 and -3.
Now we have 3(x² + 3x - 3x - 9).
Let's factor.
3(x(x+3)-3(x+3))
<u>3(x-3)(x+3)</u>
There are multiple shortcuts which you could make here, FYI:
Instead of splitting the middle, if your a value is 1, you can go straight to that step (x+number)(x+other number).
Whenever you have a difference of squares, like a²-b², that factors to (a+b)(a-b).
Y=-2x-2 it is the only one that is perfectly perpendicular and passes through (-3,4)
Answer:
The first one (A) is the answer I got!
Step-by-step explanation:
The answer will be the set of points that contains a different x value for each point.
We clearly see that for ever point in terms of CHOICE C the value of x is different.
Answer: Choice C
