Question

How do you find the intercepts, vertex and graph `f(x)=3x^2+6x-1`?

Answer 1

y-int `(0,-1)`
x-int `(-1+-2/3sqrt3,0)`
vertex `(-1,-4)`

Explanation:

to find the intercepts

∙
let x = 0, in the equation for y-intercept
`3*0^2+6*0-1`
`=3^2-1`
`=0-1`
`=-1`
`y-`int =>`(0,-1)`
∙

let y = 0, in the equation for x-intercepts
since this equation is not factorable, use uadratic fomula to solve for `x`
`x=(-b+-sqrt(b^2-4ac))/(2a)`
`3x^2+6x-1=0`
a=3
b=6
c=-1
answer:
`x=-1+-(-2)/3sqrt3`

for parabola `ax^2+bx+c` the vertex's x equals `-b/(2a)`
`x=-6/(2*3)`
`x=-1`

plug in -1 to find y value

`y=3(-1)^2+6(-1)-1`
`y=-4`

vertex is `(-1,-4)`