How do you identify the important parts of #f(x)= 3x²+x-5# to graph it?

1 Answer
Jul 24, 2018

Vertex#(-1/6, -61/12)#
y-intercept#(0,-5)#
X-intercepts #((sqrt61-1)/6, 0); (-(sqrt61-1)/6, 0)#

Explanation:

Given -

#f(x)=3x^2+x-5#

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

Vertex

#x=(-b)/(2a)=(-1)/(2xx3)=-1/6#

At #x=-1/6;y= 3(-1/6)^2+(-1/6)-5#

#y=1/12-1/6-5=(1-2-60)/12=61/12=-61/4#

Vertex#(-1/6, -61/12)#

y-intercept
At# x=0; y=3(0)^2+(0)-5=5#

y-intercept#(0,-5)

x-intercepts

#3x^2+x=5#

#x^2+1/3x=5/3#

#x^2+1/3x+1/36=5/3+1/36=(60+1)/36=61/36#

#(x+1/6)^2=61/36#

#x+1/6=+-sqrt(61/36)=+-sqrt61/6#

#x=sqrt(61/6)-1/6=(sqrt61-1)/6#

#x=-sqrt(61/36)-1/6=-(sqrt61-1)/6#

X-intercepts #((sqrt61-1)/6, 0); (-(sqrt61-1)/6, 0)#

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