SOLUTION: Use completing the square to find the zeros of the function y=9x^2-12x-33.

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Question 121503: Use completing the square to find the zeros of the function y=9x^2-12x-33.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=9+x%5E2-12+x-33 Start with the given equation



y%2B33=9+x%5E2-12+x Add 33 to both sides



y%2B33=9%28x%5E2%2B%28-4%2F3%29x%29 Factor out the leading coefficient 9



Take half of the x coefficient -4%2F3 to get -2%2F3 (ie %281%2F2%29%28-4%2F3%29=-2%2F3).


Now square -2%2F3 to get 4%2F9 (ie %28-2%2F3%29%5E2=%28-2%2F3%29%28-2%2F3%29=4%2F9)





y%2B33=9%28x%5E2%2B%28-4%2F3%29x%2B4%2F9-4%2F9%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 4%2F9 does not change the equation




y%2B33=9%28%28x-2%2F3%29%5E2-4%2F9%29 Now factor x%5E2%2B%28-4%2F3%29x%2B4%2F9 to get %28x-2%2F3%29%5E2



y%2B33=9%28x-2%2F3%29%5E2-9%284%2F9%29 Distribute



y%2B33=9%28x-2%2F3%29%5E2-4 Multiply



y=9%28x-2%2F3%29%5E2-4-33 Now add %2B33 to both sides to isolate y



y=9%28x-2%2F3%29%5E2-37 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=9, h=2%2F3, and k=-37. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=9x%5E2-12x-33 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C9x%5E2-12x-33%29 Graph of y=9x%5E2-12x-33. Notice how the vertex is (2%2F3,-37).



Notice if we graph the final equation y=9%28x-2%2F3%29%5E2-37 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C9%28x-2%2F3%29%5E2-37%29 Graph of y=9%28x-2%2F3%29%5E2-37. Notice how the vertex is also (2%2F3,-37).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.