What are the important points needed to graph #f(x)= 3x²+x-5#?

1 Answer
May 21, 2018

#x_1=(-1-sqrt61)/6#

#x_2=(-1+sqrt61)/6#

are solutions of #f(x)=0#

#y=-61/12#

is the minimum of the function

See explanations below

Explanation:

#f(x)=3x²+x-5#

When you want to study a function, what is really important are particular points of your function: essentially, when your function is equal to 0, or when it reaches a local extremum; those points are called critical points of the function: we can determine them, because they solve : #f'(x)=0#

#f'(x)=6x+1#

Trivially, #x=-1/6#, and also, around this point, #f'(x)#

is alternatively negative and positive, so we can deduce that

So : #f(-1/6)=3*(-1/6)²-1/6-5#

#=3*1/36-1/6-5#

#=1/12-2/12-60/12#

#f(-1/6)=-61/12#

is the minimum of the function.

Also, let's determine where #f(x)=0#

#3x²+x-5=0#

#Delta=b²-4ac#

#Delta=1²-4*3*(-5)#

#Delta=61#

#x=(-b+-sqrtDelta)/(2a)#

So :

#x_1=(-1-sqrt61)/6#

#x_2=(-1+sqrt61)/6#

are solutions of #f(x)=0#

\0/ Here's our answer !