Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Graphing $y=ax^2+ bx + c$ by completing the square

  1. Add and subtract the square of half the coefficent of $x$.

  2. Group the perfect square trinomial.

  3. Write the trinomial as a square of a binomial.

Rewrite $y = x^2 + 6x + 8$ into $y = a(x-h)^2 + k$.

I've tried solving this but I get a bit confused at the step where I have to "write the trinomial as a square of a binomial". Not exactly sure how to do that.

share|cite|improve this question
Welcome to Math.SE. Thank you for your question. We will be able to better answer it if you give the context of this question, including a more complete quote than the sentence fragment you cite. Also, if you've tried anything so far, please share that as well. – vadim123 Apr 29 '13 at 2:27
"Thank you for your question"? Who thanks for a question asked to him? And perhaps more interesting: why ?! – DonAntonio Apr 29 '13 at 2:31
@DonAntonio why not?! – amWhy Apr 29 '13 at 3:10
@amWhy, why why not? We can continue with for ages and it will get pretty boring after a while, and believe me: I know what is "answering" a question with another question... – DonAntonio Apr 29 '13 at 9:02
@SS' : I suggest you change the title of your question, if that is possible. You are trying to graph a quadratic function, which is a little harder than solving the corresponding quadratic equation. – Stefan Smith Apr 30 '13 at 2:40

Hint $\rm\,\ \ X^2\! + 2b\, X\! + c\ =\ \overbrace{(X^2\! + 2b\,X\! + b^2)}^{\rm complete\ \ the\ \ \color{#c00}{square}}\! -\! b^2\!+c\ =\ \overbrace{(X + b)^2}^{\rm \color{#c00}{square}} - b^2\!+c $

share|cite|improve this answer


and now try difference of squares...

In general:

$$x^2\pm ax=\left(x\pm\frac a2\right)^2-\frac{a^2}4$$

share|cite|improve this answer
It might help OP to see $x^2+6x+8=x^2+6x+9-9+8=(x+3)^2-1$. Just adding a step. – Ross Millikan Apr 29 '13 at 2:46

To complete the square for the equation:

$y = x² + 6x + 8$

Start by:

1) $x^2+6x+8$

2) $x^2+\dfrac 62x +8$ *divide the bx by 2

3) $(x^2+\dfrac 62x + (\dfrac 62)^2)+8$ *square the bx term to form a perfect square

4) $(x^2+ 3x + (36/4))+8$

5) $(x^2+3x + 9)+8-9$ *add the inverse of the c term

6) $(x^2+3x+9)-1$ *factor the perfect square

7) $(x+3)^2-1$

8) $y=(x+3)^2-1$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.