How do you find the vertex and the intercepts for y=2x^2-12x?

Apr 27, 2016

color(blue)( y_("intercept")" is at y=0")

color(blue)(x_("intercepts")-> x=0; x=4)

color(blue)("Vertex"->(x,y)=(3,-18))" "->" and is a minimum"

Explanation:

Write as:$\text{ } y = 2 {x}^{2} - 12 x + 0$

$\textcolor{b l u e}{\text{The y intercept is at y=0}}$

y =0 at x=0

$\textcolor{b r o w n}{\text{so one of the x intercepts is at x=0}}$
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write as $y = 2 \left({x}^{2} - 6 x\right)$

$\textcolor{b l u e}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \left(- 6\right) = + 3}$

color(brown)(y=2x^2-12xcolor(green)(" "->" "y_("vertex")=2(3)^2-12(3) =-18))

$\textcolor{b l u e}{{y}_{\text{vertex}} = - 18}$

$\textcolor{b l u e}{\text{Vertex} \to \left(x , y\right) = \left(3 , - 18\right)}$

$\textcolor{b r o w n}{\text{As the coefficient of "x^2" is positive then the general shape of}}$$\textcolor{b r o w n}{\text{the graph is } \cup}$ $\textcolor{b l u e}{\text{ Thus the vertex is a Minimum}}$
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${x}_{\text{intercpt}} \to y = 0$

$\implies 0 = 2 \left({x}^{2} - 6 x\right)$

Divide both sides by 2

$\implies \frac{0}{2} = \frac{2}{2} \left({x}^{2} - 6 x\right)$

But $\frac{0}{2} = 0 \text{ and } \frac{2}{2} = 1$

$\implies 0 = \left({x}^{2} - 6 x\right)$

Factor out $x$

$\implies 0 = x \left(x - 4\right)$

For y=0 ; x=0" and/or "x=+4

color(blue)(x_("intercepts")-> x=0; x=4)
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