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

May 21, 2018

${x}_{1} = \frac{- 1 - \sqrt{61}}{6}$

${x}_{2} = \frac{- 1 + \sqrt{61}}{6}$

are solutions of $f \left(x\right) = 0$

$y = - \frac{61}{12}$

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 : $f ' \left(x\right) = 0$

$f ' \left(x\right) = 6 x + 1$

Trivially, $x = - \frac{1}{6}$, and also, around this point, $f ' \left(x\right)$

is alternatively negative and positive, so we can deduce that

So : f(-1/6)=3*(-1/6)²-1/6-5

$= 3 \cdot \frac{1}{36} - \frac{1}{6} - 5$

$= \frac{1}{12} - \frac{2}{12} - \frac{60}{12}$

$f \left(- \frac{1}{6}\right) = - \frac{61}{12}$

is the minimum of the function.

Also, let's determine where $f \left(x\right) = 0$

3x²+x-5=0

Delta=b²-4ac

Delta=1²-4*3*(-5)

$\Delta = 61$

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

So :

${x}_{1} = \frac{- 1 - \sqrt{61}}{6}$

${x}_{2} = \frac{- 1 + \sqrt{61}}{6}$

are solutions of $f \left(x\right) = 0$