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

Jul 24, 2018

Vertex$\left(- \frac{1}{6} , - \frac{61}{12}\right)$
y-intercept$\left(0 , - 5\right)$
X-intercepts ((sqrt61-1)/6, 0); (-(sqrt61-1)/6, 0)

#### Explanation:

Given -

$f \left(x\right) = 3 {x}^{2} + x - 5$

$y = 3 {x}^{2} + x - 5$

Vertex

$x = \frac{- b}{2 a} = \frac{- 1}{2 \times 3} = - \frac{1}{6}$

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

$y = \frac{1}{12} - \frac{1}{6} - 5 = \frac{1 - 2 - 60}{12} = \frac{61}{12} = - \frac{61}{4}$

Vertex$\left(- \frac{1}{6} , - \frac{61}{12}\right)$

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

y-intercept(0,-5)

x-intercepts

$3 {x}^{2} + x = 5$

${x}^{2} + \frac{1}{3} x = \frac{5}{3}$

${x}^{2} + \frac{1}{3} x + \frac{1}{36} = \frac{5}{3} + \frac{1}{36} = \frac{60 + 1}{36} = \frac{61}{36}$

${\left(x + \frac{1}{6}\right)}^{2} = \frac{61}{36}$

$x + \frac{1}{6} = \pm \sqrt{\frac{61}{36}} = \pm \frac{\sqrt{61}}{6}$

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

$x = - \sqrt{\frac{61}{36}} - \frac{1}{6} = - \frac{\sqrt{61} - 1}{6}$

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