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

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Answer 1 · 2016-04-10T20:55:59

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

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

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

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

$x~~3.633" and "-0.367$

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