What is the vertex of #y=x^2+12x+26#?

Question

Hazel Anglin · Accepted Answer

The vertex is at #(-6, -10)#

You can find the vertex (turning point) by first finding the line that is the axis of symmetry.

#x = (-b)/(2a) = (-12)/(2(1)) = -6" "larr# This is the #x#-value of the vertex.

Now find #y#.

#y = x^2 +12x+26#

#y = (-6)^2 +12(-6) +26# #y= 36-72+26# #y=-10" "larr# This is the #y#-value of the vertex.

The vertex is at #(-6, -10)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ You can also find the vertex by completing the square to get the equation in vertex form: #y = a(x+b)^2 +c#

#y = x^2 +12x+26# #y= x^2 +12x color(red)(+6^2) color(red)(-6^2) +26" " color(red)((b/2)^2 =(12/2)^2)#

#y = (x+6)^2 -10#

Vertex is at #(-b, c)" "rarr (-6, -10)#

Eli Assouline · Answer

undefined To find the vertex of the quadratic function y = x^2 + 12x + 26, you can use the formula x = -b/(2a) to find the x-coordinate of the vertex, where a is the coefficient of x^2 and b is the coefficient of x. Then substitute the x-coordinate back into the equation to find the y-coordinate.

In this case:
a = 1 (coefficient of x^2)
b = 12 (coefficient of x)

x = -b/(2a) = -12/(2*1) = -6

Now substitute x = -6 into the equation:
y = (-6)^2 + 12(-6) + 26
y = 36 - 72 + 26
y = -10

So, the vertex of the quadratic function y = x^2 + 12x + 26 is (-6, -10).