What is the standard form of #y=5(x-3)^2+3#?

2 Answers
Nov 17, 2015

Standard form #-> ax^2+bx+c#
If you meant to ask: 'What is this equation when presented in standard form'? Then you have #-> y=5x^2-6x+12#

Explanation:

The standard form is #y=ax^2+bx+c#

However, if you wish to present this equation in standard form we have:

#y= 5(x^2-6x+9) +3#
#y=5x^2-6x+12#

Nov 17, 2015

#5x^2−30x+48#

Explanation:

  • Expand #(x-3)^2# in the equation:

#5(x−3)(x-3)+3#

  • Distribute the 5 to the first parenthesis:

#5(x-3#
#= (5 * x)+(5*#-#3)#
#= 5x-15#

So now you have #5x−15(x-3)+3#.

  • Now distribute the #5x# & #-15# to the next parenthesis:

#5x−15(x-3)#
#= (5x*x)+(5x*#-#3)+(#-#15*x)+(#-#15*#-#3)#
#=5x^2-15x-15x+45#
#=5x^2-30x+45#

So now you have #5x^2-30x+45+3#.

  • Finally, add the constant #(#+#3)#:

#45+3 = 48#

  • Final Answer:

#5x^2-30x+48#