write the quadratic 3x^2+9x+6 in standard form (so it is clear what the vertex of the parabola would be)

Question

write the quadratic 3x²+9x+6 in standard form (so it is clear what the vertex of the parabola would be)

Answer 1

Well 3x² + 9x + 6 is standard form.  Perhaps you meant to write it in vertex form, a(x-h)² + k, where (h,k) is the location of the vertex.  To convert from standard to vertex form, complete the square as follows:

 

1)  Factor out the coefficient of the x² term from the first two terms:

 

     3(x² + 3x) + 6

 

2)  To the terms in parentheses, add 1/2 of the coefficient of the x term, squared.

 

     3(x² + 3x + (3/2)²) + 6 = 3(x² + 3x + 9/4) + 6

 

3)  Now, since you added 3·(9/4) = 27/4 to the expression, subtract that amount from the 6 to keep it the same

 

     3(x² + 3x + 9/4) + 6 - 27/4 = 3(x² + 3x + 9/4) - 3/4

 

4)  Now (x² + 3x + 9/4) = (x + 3/2)² so we have the vertex form:

 

     3(x+3/2)² - 3/4

 

     The vertex is located at (-3/2, -3/4)