0

What are the exact steps to solving these sort of questions? For example, if I were told to sketch
y = |2/6-x| + 3
Would I sketch what is inside the modulus, then make the part of the graph that is below the x axis and reflect it into the x axis? And then apply any transformations that are not within the modulus (in this case translate it by 3 units up)?
If I could do solve this question using the method above, how would I solve this question using a hybrid approach? What does the hybrid tell me? The x-intercepts? When would I use the hybrid and how?


Notice: Undefined index: title in /home3/wmroi/public_html/merspi.com.au/qa-theme/NewMerspi/qa-theme.php on line 1251

1 Answer

0
 
Best answer

Your approach works. Using a hybrid approach would involve finding when (2/6-x) is negative, and then splitting into two arcs: (2/6-x)+3 and -(2/6-x)+3. Effectively the same of what you're describing visually.

Using the hybrid function is better if you need an algebraic form of the function rather than just a graph. Algebraic form gives you the robustness to perform many other follow-on calculations - differentiate it, solve for certain points exactly, etc.


Notice: Undefined index: title in /home3/wmroi/public_html/merspi.com.au/qa-theme/NewMerspi/qa-theme.php on line 1251
What would the hybrid function and its domains for this particular equation?
To be honest, I didn't do it because I wasn't sure you had written your question properly. When you write 2/6-x, do you mean to say (2/6)-x (which what you wrote technically means) or 2/(6-x) (which is what I think you meant, because otherwise it's a pretty trivial question). To answer your question, you should solve (2/6-x) = 0, and then on either side it will be negative or positive.