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"):
$\textcolor{g r e e n}{x = \pm \frac{\sqrt{24}}{3} - 2 \text{ Exact values}}$

$\textcolor{g r e e n}{x \approx 3.633 \text{ and "-0.367" Approx values}}$

Explanation:

Given;$\text{ } y = 3 {x}^{2} + 12 x + 4$

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

Transpose equation into vertex form
Standard form $y = a {x}^{2} + b x + c$
Vertex form $y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c - \left[{\left(\frac{b}{2}\right)}^{2}\right]$

$y = 3 {\left(x + 2\right)}^{2} + 4 - 12$

$y = 3 {\left(x + 2\right)}^{2} - 8$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~

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 $\text{ "color(brown)(y=3(x+2)^2-8)" to } \textcolor{g r e e n}{3 {\left(x + 2\right)}^{2} - 8 = 0}$

${\left(x + 2\right)}^{2} = \frac{8}{3}$

Take square roots of both sides

$x + 2 = \pm \sqrt{\frac{8}{3}}$

$x = \pm \sqrt{\frac{8}{3}} - 2 \text{ } \to \frac{\sqrt{8}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{24}}{3}$

$x = \pm \frac{\sqrt{24}}{3} - 2 \text{ }$ Exact values

$x \approx 3.633 \text{ and } - 0.367$