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

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