How do you find the axis of symmetry, and the maximum or minimum value of the function #f(x)=3(x+2)(x-2)#?

Question

Hazel Anglin · Accepted Answer

#color(green)("Vertex"->(x,y)->(0,12))#

#color(green)("Axis of symmetry "-> x=0)#

#color(green)("The vertex is a minimum")# Example: suppose we had #a^2-b^2# this is the same as #(a-b)(a+b)# Note that #(x-2)(x+2)# is of the same structure Multiply out the brackets #" "3(x^2-2^2) = 3x^2+12" "larr 3(-2)^2=3xx4=12# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Observation 1")# The #3x^2# is positive so the general shape of the graph is #uu# thus there is a #color(green)("minimum")#. If it had been #-3x^2# then the general shape would have been #nn# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Observation 2")# This is a quadratic equation so the #color(green)("axis of symmetry is the "x" coordinate of the vertex")# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Determine the y intercept")# Consider the standard form of #y=ax^2+bx+c# In the question's equation ther is no #bx# term but there is a #c# term in that #c=+12# #color(green)(y_("intercept")=c=+12)# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Observation 3") ->("Determine x vertex")# The #bx# term in #y=ax^2+bx+c# moves the graph left or right from the y axis. As there is no # bx # term the graph is symmetrical about the y-axis. #color(green)("Thus "x_("vertex")-> x=0)" "color(red)(larr" axis of symmetry")# '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Observation 4") ->("Determine y vertex")# From above the graph is symmetrical about the y-axis. Consequently y-vertex must be at the point where the graph crosses the y axis, which is: #color(green)(y_("intercept")=c=+12)" "color(red)(larr" " y_("vertex")# To prove my point; substitute #x=0# into #y=3x^2+12# #color(green)(y_("vertex")=3(0^2)+12=12)# #color(green)("Vertex"->(x,y)->(0,12))#

Eli Assouline · Answer

undefined To find the axis of symmetry of the function $ f(x) = 3(x+2)(x-2) $, use the formula: $ x = \frac{-b}{2a} $. In this function, $ a = 3 $ and $ b = 0 $, so the axis of symmetry is $ x = \frac{0}{2 \cdot 3} = 0 $.

To find the maximum or minimum value of the function, you can use the vertex form of a quadratic function: $ f(x) = a(x-h)^2 + k $, where $ (h, k) $ is the vertex. In this case, $ h = 0 $, so the maximum or minimum value occurs at $ x = 0 $. Plugging $ x = 0 $ into the function gives $ f(0) = 3(0+2)(0-2) = -12 $.

Therefore, the axis of symmetry is $ x = 0 $, and the minimum value of the function is $ -12 $.