Question

PLEASE HELP ME







put the following in vertex form:

$y=x^{2} -4x+9$


$y=3x^{2} -6x+15$


$y\geq x^{2} +5x+6$

Answer 1

Answer:

1. $y = (x -2)^{2} +7\\</p><p>2. y= 3(x^{2} -1) + 14\\</p><p>3. y \geq (x + 5)^{2} -19$

Step-by-step explanation:

1. $y = x^{2} - 4x + 9\\y = (x^{2}-4x+4 ) +9 -4 (Use formula \frac{b}{2a} to get the +4 and -4)\\y = (x -2)^{2} + 7$

2. $y = 3x^{2} -6x + 15\\y = 3(x^{2} -2x +1) + 15 - 1 (Use formula \frac{b}{2a} to get the +1 and -1)\\y = 3 (x - 1)^{2} + 14$

3. $y \geq x^{2} + 5x + 6\\y\geq (x^{2} + 5x + 25) + 6 - 25 (Use formula \frac{b}{2a} to get the +25 and -25)\\y\geq (x + 5)^{2} - 19$