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

Jan 23, 2016

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

#### Explanation:

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

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

n = ${\left(\frac{b}{2}\right)}^{2}$

n = ${\left(\frac{6}{2}\right)}^{2}$

n = 9

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

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

y = 1${\left(x + 3\right)}^{2}$ - 12

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

Exercises:

1. Convert each quadratic function from standard form to vertex form:

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

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

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

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

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

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

Good luck!