How do you find the vertex and intercepts for #f(x)=3x²+12x+4#?

1 Answer
Apr 10, 2016

#color(blue)("Vertex "->(x,y)->(-2,-8)#
#color(blue)(y_("intercept")" at "x=0 -> y=4)#

#color(blue)(x_("intercepts"):#
#color(green)(x=+-sqrt(24)/3-2" Exact values")#

#color(green)(x~~3.633" and "-0.367" Approx values")#

Explanation:

Given;#" "y=3x^2+12x+4#

#color(blue)(y_("intercept")" at "x=0 -> y=4)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine "y_("intercept") " & Vertex"#

Transpose equation into vertex form
Standard form #y=ax^2+bx+c#
Vertex form #y=a(x+b/(2a))^2+c - [(b/2)^2]#

#y=3(x+2)^2+4 -12#

#y=3(x+2)^2-8#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)(x_("vertex")->(-1)xxb/(2a)" "->" "(-1)xx(2) = -2)#

#color(blue)(y_("vertex")-> c - [(b/2)^2]" "->" "-8#

#color(blue)("Vertex "->(x,y)->(-2,-8)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine " x_("intercepts"))#

Set #" "color(brown)(y=3(x+2)^2-8)" to "color(green)( 3(x+2)^2-8 =0)#

#(x+2)^2=8/3#

Take square roots of both sides

#x+2=+-sqrt(8/3)#

#x= +-sqrt(8/3)-2" " -> sqrt(8)/sqrt(3)xxsqrt(3)/sqrt(3) = sqrt(24)/3#

#x=+-sqrt(24)/3-2" "# Exact values

#x~~3.633" and "-0.367#