Old hsc question

The y-value would have to be positive, but right now, it's sitting below, so I'm guessing that I have to do (7-8k)/4 >=0?
 
I have f(x) = 2x^2 - 8x and g(x) = x+2

find the domain and range of f(g(x)) = 2*(x+2)^2 - 8*(x+2) = 2x^2 - 8

then I thought I could make y = 0 for the domain and then make x = 0 for the range.

0 = 2x^2 - 8

2x^2 = 8

x^2 = 4

x = 2

therefore,

-2 <= x <= 2

then,

y = 2*0^2 - 8

therefore,

y = -8, but now I realize I could plug 2 in to it, but I don't think that's right.
 
You had mentioned hsc in your original post
AlonzoN said:
How'd u guess?
Click to expand...
You had mentioned hsc in your original post
I have written out (because typing up properly takes too much time) the solution to that problem. Make sure you understand each step.20210801_224916.jpg
 
20210801_224934.jpg
And here's an alternative way to do the same problem. Again make sure you u derstand what's going on.
 
No the domain is the set of all of the numbers that you can put in for x in the function. Is there any value of x that you can't put into the function \(\displaystyle 2x^2-8\)?
 
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