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

Question

Eva Adamson · Accepted Answer

#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"# Write as:#" "y=2x^2-12x +0# #color(blue)("The y intercept is at y=0")# y =0 at x=0 #color(brown)("so one of the x intercepts is at x=0")# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ write as #y=2(x^2-6x)# #color(blue)(x_("vertex") = (-1/2)xx(-6) = +3)# #color(brown)(y=2x^2-12xcolor(green)(" "->" "y_("vertex")=2(3)^2-12(3) =-18))# #color(blue)(y_("vertex")=-18)# #color(blue)("Vertex"->(x,y)=(3,-18))# #color(brown)("As the coefficient of "x^2" is positive then the general shape of")##color(brown)("the graph is " uu)# #color(blue)(" Thus the vertex is a Minimum")# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #x_("intercpt")->y=0# # => 0=2(x^2-6x)# Divide both sides by 2 #=>0/2=2/2(x^2-6x)# But #0/2=0" and "2/2=1# #=>0=(x^2-6x)# Factor out #x# #=>0=x(x-4)# For #y=0 ; x=0" and/or "x=+4# #color(blue)(x_("intercepts")-> x=0; x=4)# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Eva Adamson · Answer

undefined To find the vertex of the quadratic function $ y = 2x^2 - 12x $, you can use the formula for the x-coordinate of the vertex: $ x = \frac{-b}{2a} $, where $ a $ is the coefficient of the $ x^2 $ term and $ b $ is the coefficient of the $ x $ term.

For the given function, $ a = 2 $ and $ b = -12 $. Substituting these values into the formula, you get:

$$ x = \frac{-(-12)}{2(2)} = \frac{12}{4} = 3 $$

To find the y-coordinate of the vertex, substitute $ x = 3 $ into the function:

$$ y = 2(3)^2 - 12(3) = 2(9) - 36 = 18 - 36 = -18 $$

So, the vertex of the parabola is $ (3, -18) $.

To find the x-intercepts, set $ y = 0 $ and solve for $ x $:

$$ 2x^2 - 12x = 0 $$
$$ 2x(x - 6) = 0 $$

This equation is satisfied when either $ 2x = 0 $ or $ x - 6 = 0 $.

So, the x-intercepts are $ x = 0 $ and $ x = 6 $.

To find the y-intercept, substitute $ x = 0 $ into the function:

$$ y = 2(0)^2 - 12(0) = 0 $$

So, the y-intercept is $ (0, 0) $.