What is the vertex form of $y=x^2 +6x -3$ ?

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Answer 1 · 2016-01-23T23:01:01

To convert to vertex form, you must complete the square.

y = $x^2$ + 6x - 3

y = 1( $x^2$ + 6x + n) - 3

n = $(b/2)^2$

n = $(6/2)^2$

n = 9

y = 1( $x^2$ + 6x + 9 - 9) - 3

y = 1( $x^2$ + 6x + 9) -9 - 3

y = 1 $(x + 3)^2$ - 12

So, the vertex form of y = $x^2$ + 6x - 3 is y = $(x + 3)^2$ - 12.

Exercises:

a) y = $x^2$ - 12x + 17

b) y = $-3x^2$ + 18x - 14

c) y = $5x^2$ - 11x - 19

Solve for x by completing the square. Leave any non-integer answers in radical form.

a) $2x^2$ - 16x + 7 = 0

b) $3x^2$ - 11x + 15 = 0

Good luck!

What is the vertex form of y=x^2 +6x -3 y=x^2 +6x -3 ?