The y intercept is 17. You can see it clearly in the equation.
Answer:
(x + 9)(x + 3)
Step-by-step explanation:
Factor 27 so that the factors, when combined, will equal 12:
x² + 12x + 27
x 9
x 3
(x + 9)(x + 3) is your answer.
Check: Use the FOIL method. Multiply the first two terms, the outside terms, the inside terms, and then the last two terms. Combine like terms:
(x + 9)(x + 3) = x² + 3x + 9x + 27 = x² + 12x + 27 (√)
~
An absolute value is positive value of any value. So the abs value of -28 is 28. The abs value of 67 is 67. Makes sense?
If it were |27-3| for example, treat the inside of a abs as parenthesis, so you must complete PEMDAS inside of it to reduce the equation to |24|, unless you wanted it to become |27| - |3|.
For functions, this becomes slightly different and more difficult, especially when adding a variable such as x. Look below for a sample equation.
|2x-3|=1
This equation will actually have (and most others) 2 solutions for x. To find these, you’ll need to multiply the inside of the abs by -1 for one equation, and leave it as it is for the other!
2x-3=1 -(2x-3)=1
Now you have to solve BOTH equations to get your correct x-value answers.
For the first listed equation:
2x=4
x=2
For the second listed equation:
-2x+3=1
-2x=-2
x=-1
So you get the x-values -1 and 2 which both make the parent function true!
Solve for y...
2xy=1
y=1/(2x)
So this is a function as each x value (with the exception of x=0) produces just one value for y. However, technically there is a vertical asymptote about the vertical line x=0. So there is a discontinuity as x=0.
I say this technically, because most would say that this function is true even with the discontinuity (as they do with say y=tana), but technically the discontinuity makes this not a function because another requirement of a function is for their to be an actual output value for each input value. In this case division by zero when x=0 is undefined and cannot be included as part of a function. So if you are picky about being absolute, 2xy=1 is really two functions. One with a domain of (-oo,0) and another with a domain of (0,+oo).
