Given
color(white)("XXX")y=3x^2+12x-6XXXy=3x2+12x−6
This can be converted into vertex form as follows:
rArr y=3(x^2+4x-2)⇒y=3(x2+4x−2)
rArr y=3[x^2+4x-2]⇒y=3[x2+4x−2]
rArr y = 3[(x^2+4xcolor(blue)(+4))-2color(blue)(-4)]⇒y=3[(x2+4x+4)−2−4]
rArr y=3[(x+2)^2-6]⇒y=3[(x+2)2−6]
rArr y= 3(x+2)^2-18⇒y=3(x+2)2−18
rArr y= 3(x-color(red)(""(-2)))^2+color(blue)(""(-18))⇒y=3(x−(−2))2+(−18)
which is the vertex form with vertex at (color(red)(-2),color(blue)(-18))(−2,−18)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The y-intercept is the value of yy when x=color(magenta)0x=0.
Substituting color(magenta)00 for xx in the original equation
color(white)("XXX")y=3 * color(magenta)0^2+12 * color(magenta) 0 -6=-6XXXy=3⋅02+12⋅0−6=−6
So the y-intercept is at y=-6y=−6 [or if you prefer at (0,-6)(0,−6)]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The x-intercepts are the values of xx for points when y=0y=0
Using the form derived while determining the vertex form: y=3(x+2)^2-18y=3(x+2)2−18
with y=color(green)0y=0
color(white)("XXX")color(green)0=3(x+2)^2-18XXX0=3(x+2)2−18
color(white)("XXX")rArr color(green)0=(x+2)^2-6XXX⇒0=(x+2)2−6
color(white)("XXX")rArr (x+2)^2=6XXX⇒(x+2)2=6
color(white)("XXX")rArr x+2=+-sqrt(6)XXX⇒x+2=±√6
color(white)("XXX")rArr x=-2+-sqrt(6)XXX⇒x=−2±√6
So the x-intercepts are at x=-2-sqrt(6)x=−2−√6 and x=-2+sqrt(6)x=−2+√6 [or if you prefer at (-2-sqrt(6),0)(−2−√6,0) and (-2+sqrt(6),0)(−2+√6,0)]
