SOLUTION: Make a graph of y=6x^2-4x-3 and find the vertex, y-intercept, extra points. I don't understand how I can find the vertex, y-intercept and extra points.

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Question 437124: Make a graph of y=6x^2-4x-3 and find the vertex, y-intercept, extra points.
I don't understand how I can find the vertex, y-intercept and extra points.
Found 2 solutions by mananth, Alan3354:
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
6x^2-4x-3
vertex = x=-b/2a
vertex y= c-b^2/4a
x= -(-4)/2*6= 0.33
y= -3-(16/24)= -3.6666
Vertex ( 0.33,-3.6666)
xintercepts
y=0
6x^2-4x-3=0
Find the roots of the equation by quadratic formula

a= 6 b= -4 c= -3

b^2-4ac=16-72
b^2-4ac=88 , = 9.38

x1=(4+9.38)/12
x1=1.12
x2=(4-9.38)/12
x2= -0.45
(-0.45,0) (1.12,0) x intercepts
......
y intercepts plug x=0
6x^2-4x-3
-3 is the y intercept

Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
Make a graph of y=6x^2-4x-3 and find the vertex, y-intercept, extra points.
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For the y-intercept, set x = 0
--> y = -3, or (0,-3)
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For the vertex, find the line of symmetry
It's x = -b/2a
x = -(-4)/(2*6) = 1/3
The vertex is (1/3,f(1/3))
Sub 1/3 for x
--> 6*(1/9) - 4/3 - 3
= - 11/3
Vertex at (1/3,-11/3)
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For "extra points", pick values for x and find y
x = 1, y = -1 --> (1,-1)
x = 2, y = 13 --> (2,13) etc