What is the vertex form of #y=2x^2+11x+12#?

Question

Hazel Anglin · Accepted Answer

Yhe vertex form is #y=2(x+11/4)^2-25/8#

To find the vertex form, you complete the square

#y=2x^2+11x+12#

#y=2(x^2+11/2x)+12#

#y=2(x^2+11/2x+121/16)+12-121/8#

#y=2(x+11/4)^2-25/8#

The vertex is #=(-11/4, -25/8)#

The symmetry line is #x=-11/4#

graph{(y-(2x^2+11x+12))(y-1000(x+11/4))=0 [-9.7, 2.79, -4.665, 1.58]}

Genesis Allen · Answer

#color(blue)(y=2(x+11/4)^2-25/8)#

Consider the standardised form of #y=ax^2+bx+c#

The vertex form is: #y=a(x+b/(2a))^2+k+c#

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#color(brown)("Additional note about the method")#
By rewriting the equation in this form you introduce an error. Let me explain.

Multiply out the bracket in #y=a(x+b/(2a))^2+c# and you get:

#y=a[x^2+(2xb)/(2a)+(b/(2a))^2]+c#

#color(green)(y=ax^2+bx+color(red)(a(b/(2a))^2)+c)#

the #color(red)(a(b/(2a))^2)# is not in the original equation so it is the error. Thus we need to 'get rid' of it. By introducing the correction factor of #k# and setting #color(red)(a(b/(2a))^2+k=0)# we 'force' the vertex form back into the value of the original equation.
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Given:#" "y=ax^2+bx+c" "->" "y=2x^2+11x+12#

#y=a(x+b/(2a))^2+k+c" "->" "y=2(x+11/4)^2+k+12#

But:
#a(b/(2a))^2+k=0" "->" "2(11/4)^2+k=0#

#=>k=-121/8#

So by substitution we have:
#y=a(x+b/(2a))^2+k+c" "->y=2(x+11/4)^2-121/8+12#

#color(blue)(y=2(x+11/4)^2-25/8)#
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The two equation have been plotted to show that they produce the same curve. One is thicker than the other so that they can both be seen.

Eva Abramson · Answer

undefined The vertex form of the quadratic function y = 2x^2 + 11x + 12 is y = 2(x + 11/4)^2 - 37/8.