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

2 Answers
Tony B
Nov 17, 2015

Standard form -> ax^2+bx+c→ax2+bx+c
If you meant to ask: 'What is this equation when presented in standard form'? Then you have -> y=5x^2-6x+12→y=5x2−6x+12

Explanation:

The standard form is y=ax^2+bx+cy=ax2+bx+c

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

y= 5(x^2-6x+9) +3y=5(x2−6x+9)+3
y=5x^2-6x+12y=5x2−6x+12

Janae
Nov 17, 2015

5x^2−30x+485x2−30x+48

Explanation:

  • Expand (x-3)^2(x−3)2 in the equation:

5(x−3)(x-3)+35(x−3)(x−3)+3

  • Distribute the 5 to the first parenthesis:

5(x-35(x−3
= (5 * x)+(5*=(5⋅x)+(5⋅-3)3)
= 5x-15=5x−15

So now you have 5x−15(x-3)+35x−15(x−3)+3.

  • Now distribute the 5x5x & -15−15 to the next parenthesis:

5x−15(x-3)5x−15(x−3)
= (5x*x)+(5x*=(5x⋅x)+(5x⋅-3)+(3)+(-15*x)+(15⋅x)+(-15*15⋅-3)3)
=5x^2-15x-15x+45=5x2−15x−15x+45
=5x^2-30x+45=5x2−30x+45

So now you have 5x^2-30x+45+35x2−30x+45+3.

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

45+3 = 4845+3=48

  • Final Answer:

5x^2-30x+485x2−30x+48

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