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

1 Answer
Kiana S · Amory W.
Nov 22, 2017

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)#

enter image source here

Related questions
See all questions in Graphing Quadratic Functions
Impact of this question
7646 views around the world
You can reuse this answer
Creative Commons License