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

2 Answers
Tony B
Jan 18, 2016

Vertex# ->(x.y)->(+1,?)#
Having shown you a cool trick to get the value of #x# I will let you finish it to find #y#

Explanation:

#color(blue)("To find "x_("vertex"))#

Given: #color(white)("xx")y=-3x^2+6x-2color(white)(.)# .......(1)

Write as: #-3(x^2color(brown)(-2)x)-2#

Now consider the #color(brown)(-2)" from the "-2x" inside the brackets"#

Multiply this value by #(-1/2)#

#(-1/2) xx (-2)=+1#

#color(blue)(x_("vertex")=+1)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Compare this to the graph:
enter image source here

Ok! You have now found the value for #x=1#

Substitute this back into equation (1) to find the value for #y#
I will let you do that!

Nghi N.
Jan 18, 2016

Vertex (1, 1)
y = -2
#x = (3 +- sqrt3)/3#

Explanation:

x-coordinate of vertex:
#x = -b/(2a) = -6/-6 = 1#
y-coordinate of vertex:
#f(1) = -3 + 6 - 2 = 1.#
Vertex (1, 1)
To find y-intercept, make x = 0 --> y = -2
To find x-intercepts, make y = 0 and solve:
#f(x) = -3x^2 + 6x - 2 = 0#
#D = d^2 = b^2 - 4ac = 36 - 24 = 12# --> #d = +- 2sqrt3#
#x = -b/(2a) +- d/(2a) = 1 +- sqrt3/3 = (3 +- sqrt3)/3#

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