Question

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

Answer 1

`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 !