What is the vertex form of #2y=10x^2+7x-3#?

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Answer 1 · 2018-05-05T13:08:19

#color(blue)(y=5(x+7/20)^2-169/80)#

#2y=10x^2+7x-3#

Divide by 2:

#y=5x^2+7/2x-3/2#

We now have the form:

#color(red)(y=ax^2+bx+c)#

We need the form:

#color(red)(y=a(x-h)^2+k)#

Where:

#bba color(white)(8888)# is the coefficient of #x^2#

#bbh color(white)(8888)# is the axis of symmetry.

#bbk color(white)(8888)# is the maximum or minimum value of the function.

It can be shown that:

#h=-b/(2a)color(white)(8888)# and # color(white)(8888)k=f(h)#

#:.#

#h=-(7/2)/(2(5))=-7/20#

#k=f(h)=5(-7/20)^2+7/2(-7/20)-3/2#

# color(white)(8888)=245/400-49/40-3/2#

# color(white)(8888)= 49/80-49/40-3/2#

# color(white)(8888) =(49-98-120)/80=-169/80#

Vertex form:

#y=5(x+7/20)^2-169/80#