Use the Completing the Square method to find the vertex form of the quadratic function y = x^2 + 7x + 12?

Equation: x^2+7x+12
(x^2+7 x)+12
(x^2+7 x+(7/2)^2)+12-1(7/2)^2
(x+7/2)^2+12-49/4
(x+7/2)^2-1/4
y=(x+7/2)^2-1/4
vertex is (-7/2, -1/4)

y = x² + 5x + 12 y - 12 = x² + 5x The coefficient of the x term is 5. comprehensive the sq. employing (5/2)² y - 12 + (5/2)² = x² + 5x + (5/2)² y - 12 + (25/4) = (x+5/2)² y + 13/4 = (x+5/2)² that's the equation of an up-beginning off parabola with vertex at (-5/2,-13/4)

a=1 b=7 c=12

1st, find the axis of symmetry (aos)

x=-b/2a
x= -(7)/2(1)
x=-7/2
aos= -7/2

Plug in x in the quadratic equation
(-7/2)^2+7(-7/2)+12= -1/4

Vertex= (-7/2, -1/4)

y = (x = (7 / 2))^2 + 12 - (49 / 4)
y = (x + (7 / 2)^2 - (1 / 4)

y = ( x² + 7x + 49/4) + 12 - 49/4

y = (x + 7/2)² - 1/4

Vertex V (- 7/2 , -1/4)

x^2 + 7x + 12 = y
x^2 + 7x = y - 12
x^2 + 7x/2 + 7x/2 = y - 12
x^2 + 7x/2 + 7x/2 + 49/4 = y - 12 + 49/4
(x^2 + 7x/2) + (7x/2 + 49/4) = (4y - 48)/4 + 49/4
x(x + 7/2) + 7/2(x + 7/2) = (4y - 48 + 49)/4
(x + 7/2)(x + 7/2) = (4y + 1)/4
(x + 7/2)^2 = 4y/4 + 1/4
(x + 7/2)^2 = y + 1/4

∴ the vertex is (-7/2 , -1/4)

Hi,

y = x² + 7x + 12

y = x² + 7x + __ + 12
½(7) = 3.5 and 3.5² = 12.25

y = (x² + 7x + 12.25) + 12 - 12.25

y = (x + 3.5)² - .25 <==ANSWER

I hope that helps!! :-)

Merry Christmas!!

y = x^2 + 7x + 12
y - 12 = x^2 +7x
y -12 + 49/4 = x^2 + 7x + 49/4
y +1/4 = (x+7/2)^2
so vertex V is at (-7/2, -1/4) (answer)

y = (x+3.5)^2-2.5

(x+4)(x+3)