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

1 Answer
Aug 12, 2017

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

Explanation:

#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 vertex form of the quadratic equation is:

#y=2(x+7/4)^2-25/8#

graph{y=2x^2+7x+3 [-10, 10, -5, 5]}