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!