Question 927730: Hello....I have 2 questions
1) y=x^2-6x+4 use quadratic formula to find the values of zeros.
Can someone help me with this? I believe I have found the vertex (3,-5) hoping this is correct?
2) Find the y and x intercepts of 2x^2-2x-12
Thank you!!
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! 1) y = x^2 -6x +4
using quadratic formula x = (-b + or - square root(b^2-4ac)) / 2a
x = (6 + square root(6^2 -4*1*4)) / (2*1) = 5.236067977
x = (6 - square root(6^2 -4*1*4)) / (2*1) = 0.763932023
x coordinate of vertex is -b/2a = 6/2 = 3
substitute for x in quadratic equation
y = 3^2 -6*3 +4 = -5
vertex is (3, -5)
2) y = 2x^2 -2x -12
if x = 0, then y = -12 which is the y intercept
to find x intercepts, set quadratic = 0
2x^2 -2x -12 = 0
divide both sides of = by 2
x^2 -x -6 = 0
this factors into
(x-3)*(x+2) = 0
x intercepts are 3 and -2
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