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# Algebra/Find the vertex (Algebra II)

Question
Find the Vertex of (x-3)2+5?

Solve by completing the square:
x2+2x-15 = 0?

and

Write the Quadratic function in Vertex Form:
x2+4x+5?

I take it the 2 at the end is really for a square, as in (x-3)2 is (x-3)².

If y = (x-3)² + 5, the minimum is found at x=3, and that is the vertex.
Putting in x=3 makes the value in the parenthesis 0, so that answer is y=5.
This makes the vertex at the point (3,5).

For x²+2x-15, the first thing to look at is the factors of 15.
Since 15 = 3*5, the numbers to try are 3 and 5.
Since -15 is negative, one of them needs to be negative.
Since the +2 is positive, the larger of the two needs to be positive.
This means to try (x+5)(x-3).
This gives x² + 5x - 3x -15.
Combining the x terms gives x² + 2x - 15, and that's what we're looking for.

Writing x² + 4x + 5 in quadratic form means writing it as (x+a)² + b.
To find a, note that it is half the x factor.
Since the x factor is 4, that makes a 2.
Since we have (x+2)², that works out to x² + 2x + 4.
Now that 4 added on must be subtracted from the 5 we have already, and 5-4=1.
This gives (x+2)² + 1.
The vertex can then be seen to be (-2,1).

Algebra

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#### Scott A Wilson

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