What is the vertex form of #y= 2x^2 + 8x+4 #?

Question

What is the vertex form of #y= 2x^2 + 8x+4   #?

Hazel Anglin · Accepted Answer

#y=2(x-(-2))^2+(-4)#

The general vertex form is #color(white)("XXX")y=color(green)m(x-color(red)a)^2+color(blue)b# with vertex at #(color(red)a,color(blue)b)#

Given #color(white)("XXX")y=2x^2+8x+4#

Extract the #color(green)m# factor (in this case #color(green)2# from the first two terms #color(white)("XXX")y=color(green)2(x^2+4x)+4#

Supposing that the #(x^2+4x)# are part of a squared binomial #(x-color(red)a)^2=(x^2-2color(red)ax+color(red)a^2)# then #-2color(red)ax# must be equal to #4x# #rarr color(red)a=4/(-2)=color(red)(-2)# and in order to "complete the square" an extra #color(red)a^2=(-2)^2=color(magenta)4# will need to be added inside the brackets to the #(x^2+4x)# we already have. Note that if we insert this #color(magenta)(+4)# because of the factor #color(green)m=color(green)2# we will really be adding #color(green)2 xx color(magenta)4=color(brown)8# to the expression.

To maintain equality if we add #color(brown)8# we will also need to subtract it: #color(white)("XXX")y=color(green)2(x^2+4x+color(magenta)4)+4color(brown)(-8)# Simplifying #color(white)("XXX")y=color(green)2(x+2)^2-4#

Adjusting to match the sign requirements of the vertex form: #color(white)("XXX")y=color(green)2(x-color(red)(""(-2)))^2+color(blue)(""(-4))#

This result is corroborated by the graph of the original equation below: graph{2x^2+8x+4 [-8.386, 2.71, -5.243, 0.304]}

Ethan Adkins · Answer

undefined The vertex form of the given quadratic function is y = 2(x + 2)^2 - 4.

Kennedy Adams · Answer

undefined The vertex form of the quadratic function $ y = 2x^2 + 8x + 4 $ is:

$$ y = 2(x - h)^2 + k $$

To find the vertex form, we complete the square.

1. Factor out the leading coefficient from the $ x^2 $ and $ x $ terms:
   $$ y = 2(x^2 + 4x) + 4 $$

2. Complete the square inside the parentheses by adding and subtracting $(4/2)^2 = 4$:
   $$ y = 2(x^2 + 4x + 4 - 4) + 4 $$

3. Rewrite the expression inside the parentheses as a perfect square:
   $$ y = 2((x + 2)^2 - 4) + 4 $$

4. Distribute the 2 outside the parentheses:
   $$ y = 2(x + 2)^2 - 8 + 4 $$

5. Combine like terms:
   $$ y = 2(x + 2)^2 - 4 $$

Therefore, the vertex form of the quadratic function $ y = 2x^2 + 8x + 4 $ is $ y = 2(x + 2)^2 - 4 $.