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.

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) (Show Source): You can put this solution on YOUR website!
the standard form
change it to vertex form using completing the square

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

Start with the given equation

Subtract from both sides

Factor out the leading coefficient

Take half of the x coefficient to get (ie ).

Now square to get (ie )

Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of does not change the equation

Now factor to get

Distribute

Multiply

Now add to both sides to isolate y

Combine like terms

Now the quadratic is in vertex form where , , and . Remember (h,k) is the vertex and "a" is the stretch/compression factor.

Check:

Notice if we graph the original equation we get:

Graph of . Notice how the vertex is (,).

Notice if we graph the final equation we get:

Graph of . Notice how the vertex is also (,).

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 (,)

here is better graph

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