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Question 3900: What is the slope of the line: x/3 - (- y/5) = 1
What is the domain for g(x) = (6 + x)(4 + x)
What is the domain for f(x) = 2(x - 5)^2 - 8
What is the range for f(x) = 9 - 8x - x^2
Answer by longjonsilver(2297)   (Show Source):
You can put this solution on YOUR website! 1. x/3 - (- y/5) = 1
--> x/3 + y/5 = 1
Multiply every term by 15, to get rid of the fractions:
15x/3 + 15y/5 = 15
5x + 3y = 15
3y = -5x + 15
y = (-5/3)x + 15/3
--> y = (-5/3)x + 5
In this form, the gradient is the coefficient of the x-term...ie gradient = -5/3
2. g(x) = (6 + x)(4 + x)
this is a quadratic. As such, any x-value is possible to put into the equation. The domain is just the posh word for the "numbers you put into the equation"...so domain is "any real number", written as xeR.
3. f(x) = 2(x - 5)^2 - 8
Again...which x-values can we put into the equation? Answer? any...so domain is xeR again. Easy.
4. f(x) = 9 - 8x - x^2. Now the range is tricker..this is the possible answers we get out when we put in the domain values (the x-values).
Being a quadratic, the domain is any value of x. So what are the possible answers? I shall leave that up to you. Use calculus to find the turning point.
jon.
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