What is the vertex form of #y=2x^2+7x+3 #?

Question

What is the vertex form of #y=2x^2+7x+3     #?

Eli Assouline · Accepted Answer

The vertex form is #y=2(x+7/4)^2-25/8#.

#y=2x^2+7x+3# is a quadratic equation in standard form: #y=ax^2+bx+c#, where #a=2#, #b=7#, and #c=3#. The vertex form is #y=a(x-h)^2+k#, where #(h,k)# is the vertex. In order to determine #h# from the standard form, use this formula: #h=x=(-b)/(2a)# #h=x=(-7)/(2*2)# #h=x=-7/4# To determine #k#, substitute the value of #h# for #x# and solve. #f(h)=y=k# Substitute #-7/4# for #x# and solve. #k=2(-7/4)^2+7(-7/4)+3# #k=2(49/16)-49/4+3# #k=98/16-49/4+3# Divide #98/16# by #color(teal)(2/2# #k=(98-:color(teal)(2))/(16-:color(teal)(2))-49/4+3# Simplify. #k=49/8-49/4+3# The least common denominator is #8#. Multiply #49/4# and #3# by equivalent fractions to give them a denominator of #8#. #k=49/8-49/4xxcolor(red)(2/2)+3xxcolor(blue)(8/8# #k=49/8-98/8+24/8# #k=-25/8# The quadratic equation has the following vertex form: #y=2(x+7/4)^2-25/8# graph{y=2x^2 + 7x + 3 [-10, 10, -5, 5]}

Savannah Anderson · Answer

undefined The vertex form of the quadratic equation y = 2x^2 + 7x + 3 is y = 2(x + 7/4)^2 - 25/8.