You do this by taking the coefficient of x, which is -6, and take half of it, and square, which is +9. The trick is to add +9 and subtract 9 on the right side, in order to do what needs to be done, but keep the equation the same.

Question 860744: How do I use the formula y=a(x-p)^2+q to determine the equation of the graph not given any points Answer by ewatrrr(24785) ( Show Source ): You can put this solution on YOUR website!

Question 1209496: Express x^2 + 6kx + 144 in the form of (x+p)^2 + q. Find the range of values of k suck that the x^2 + 6kx + 144 is positive for all values of x. [the answer for (a) is solved already, I need the second part please.] [the answer for (b) is -4 Thank you! Answer by ikleyn(51916) (Show Source):

Question 431533: x^2+6x-2=(x+p)^2+q find values of p and q please help the answer is p=3 and q=11, but tried to workd out but dont know how thanks Answer by sudhanshu_kmr(1152) ( Show Source ):

The equation 2x^2-5x=-12 is rewritten in the form of 2(x-p)^2+q=0. So we have the identity Substitute x = 0 Substitute x=p Equate the two expressions for q ≡ ≡ ≡ or But p cannot be 0, since ≡ ≡ cannot be an identity, since there is no x term on the left So (p,q) = But the first Edwin

y=a(x+p)2+q Is that all you have? If a is positive, the vertex is a minimum point. If a is negative, the vertex is a maximum point. The vertex is (-p,q). The x-axis intercepts, if the graph actually has any, would be found using -----the zeros or roots

Question 962474: Use the method of completing the square to transform the quadratic equation into the equation form (x + p)2 = q. 1 + x + 2x2 = 0 Answer by Theo(13342) ( Show Source ):

Question 967930: Use the method of completing the square to transform the quadratic equation into the equation form (x – p)2 = q. -12 + -5x2 + 2x4 = 0 Answer by josgarithmetic(39581) ( Show Source ):

The easiest way to graph it is to turn it into standard form [y=a(x-p)^2+q], by completing the square. y=x^2 +5x+6 Halve the b term (b=5 in this equation), then square it. (5/2)^2 = 25/4 Add this number to the first two terms, but then subtract it from the end (so that you are only adding zero to the equation). y = x^2 +5x +25/4 +6 -25/4

y=(x+2)^2 -1 Now the equation is in standard form, y=a(x-p)^2 + q. The vertex is at (p,q)=(-2,-1). You can find the y=intercept by setting x=0 and solving for y. It's easiest to do this with the general form: y=x^2+4x+3 y = 0+)+3 y=3 The y-intercept is y=3. To find the x-intercepts, factor, set y=0 and solve for x. You have to solve by ...