How do you graph #y=2x^2+12x+16#?

1 Answer
Mar 19, 2018

There is a procedure to graph funcion.

Explanation:

  1. Define the demain and codomain: polynoms are defined everywhere in the real plane (complex too).
    #y=2*x^2+12*x+16#
  2. Find the intersection between function and x-axes: solve the equation #2*x^2+12*x+16=0# | :2
    #x^2+6*x+8=0#
    Remember: #(a+x)*(b+x)=x^2+(a+b)*x+a*b#
    The solution is: #x^2+6*x+8=(x-4)*(x-2)=0# #x_1=4, x_2=2#
    [black points]
  3. Calculate the first derivative: #y'=4*x+12#
  4. Calculate #y'=0#:
    #4*x+12=0#, #x=-3#
    here #(-3,y(-3))=(-3,-2)# blue point the slope change. #(-infty,-3)# the slope is negative (the value of the function fall), #(-3,+infty)# the slope is positive (the value of the function increase).
  5. Calculate the second derivative: #y''=4#
    if #y''>0 # the function is convex (is smiling)
    if #y''<0 # the function is concave (is sad)
    #y''=4 >0 rArr# is convex

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