SOLUTION: What is the domain and range of the parabola -14x^2+36x-30=0 ?

Algebra.Com
Question 2078: What is the domain and range of the parabola -14x^2+36x-30=0 ?
Answer by matthew_sessoms(39)   (Show Source): You can put this solution on YOUR website!
You can "complete the square" (CTS) this equation or you can graph it. I will use the CTS method and then graph it to prove my statements.


Factor out the -14 from the x terms only
-14(x^2 + 18/7)-30=0


Now we can "complete the square." Take half of 18/7 (which is 9/7) and then square it (which is 81/49). Then take 81/49 and multiply it by -14 (which is -162/7) and change its sign and add it to the -30.
-14(x^2 + 18/7 + 81/49) + 162/7 - 30
-14(x + 9/7)^2 - 48/7
This is in the form of y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola.


Ok. We have now completed the square. **Remember this** EVERY quadratic equation has a domain of (-infinity, infinity )


The range is determined from the second coordinate of the vertex (k, which is -48/7). However, you must look at a (which -14) first. If a < 0 (like yours, -14) then the range is always (-infinity, k].


Therefore,
domain is (-infinity, infinity)
range is (-infinity, -48/7)



Here is the graph to prove my statements.


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