SOLUTION: I am unsure of how to do this question. Algebraically determine the vertex of the parabola from the standard form (y= -6x^2 + 48x + 9) of the function using another method.

Algebra ->  Linear-equations -> SOLUTION: I am unsure of how to do this question. Algebraically determine the vertex of the parabola from the standard form (y= -6x^2 + 48x + 9) of the function using another method.       Log On


   

Question 698656: I am unsure of how to do this question.
Algebraically determine the vertex of the parabola from the standard form (y= -6x^2 + 48x + 9) of the function using another method.
The first part of the question was to change it to vertex form using completing the square, I did that but I'm not sure what to do above.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
the standard form y=+-6x%5E2+%2B+48x+%2B+9
change it to vertex form using completing the square

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


y=-6+x%5E2%2B48+x%2B9 Start with the given equation



y-9=-6+x%5E2%2B48+x Subtract 9 from both sides



y-9=-6%28x%5E2-8x%29 Factor out the leading coefficient -6



Take half of the x coefficient -8 to get -4 (ie %281%2F2%29%28-8%29=-4).


Now square -4 to get 16 (ie %28-4%29%5E2=%28-4%29%28-4%29=16)





y-9=-6%28x%5E2-8x%2B16-16%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 16 does not change the equation




y-9=-6%28%28x-4%29%5E2-16%29 Now factor x%5E2-8x%2B16 to get %28x-4%29%5E2



y-9=-6%28x-4%29%5E2%2B6%2816%29 Distribute



y-9=-6%28x-4%29%5E2%2B96 Multiply



y=-6%28x-4%29%5E2%2B96%2B9 Now add 9 to both sides to isolate y



y=-6%28x-4%29%5E2%2B105 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=-6, h=4, and k=105. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=-6x%5E2%2B48x%2B9 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C-6x%5E2%2B48x%2B9%29 Graph of y=-6x%5E2%2B48x%2B9. Notice how the vertex is (4,105).



Notice if we graph the final equation y=-6%28x-4%29%5E2%2B105 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C-6%28x-4%29%5E2%2B105%29 Graph of y=-6%28x-4%29%5E2%2B105. Notice how the vertex is also (4,105).



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






the vertex of the parabola is at (4,105)

here is better graph

+graph%28+600%2C600%2C+-6%2C+10%2C+-5%2C+120%2C+-6x%5E2+%2B+48x+%2B+9%29+