What is the the vertex of #y =3x^2+4x-18 #?

2 Answers
Tony B
Feb 26, 2016

#x_("vertex")=-2/3" "#I will let the reader find #""y_("vertex")#

Explanation:

Given:#" "y=3x^2+4x-18" "#..................................(1)

Write as:#" "y=3(x^2+4/3x)-18#

Using the #+4/3" from "(x^2+4/3x)#

#(-1/2)xx4/3 =-4/6=-2/3#

#color(blue)(x_("vertex") = -2/3)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#-2/3" " =" " -0.6666bar6" "=" "-0.6667# to 4 decimal places

#color(brown)("All you have to do now is substitute "x=-2/3" into")##color(brown)("equation (1) to find "y_("vertex"))#

Tony B

P dilip_k
Feb 26, 2016

May be done as follows

Explanation:

The given equation is
#y=3x^2+4x-18#
#=>y=3[x^2+2x(2/3)+(2/3)^2-(2/3)^2 -6]#

#=>y=3[(x+2/3)^2-(2/3)^2 -6]#
#=>y=3[(x+2/3)^2-4/9- 6]#
#=>y=3[(x+2/3)^2-58/9 ]#
#=>y=3(x+2/3)^2-58/9*3 #
#=>y+58/3=3(x+2/3)^2 #
putting ,#y+58/3=Y and x+2/3=X # we have
new equation
#Y =3X^2#,which has coordinate of vertex (0,0)
So putting X=0 and Y=0 in the above relation
we get
#x=-2/3#
and #y=-58/3 =-19 1/3#

so the actual coordinate of vertex is # (-2/3,-19 1/3)#

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