There exist constants a,h,k and such that 3x^2+12x+4=a(x-h)^2+k
for all real numbers x Enter the ordered triple (a,h,k)
Find the vertex of the graph of the equation x-y^2+8y=13
find the vertex of the graph of the equation y=-2x^2+8x-15
1) Rewrite 3x2 + 12x + 4 into the form a(x - h)2 + k
3x2 + 12x + 4
Separate the x-terms from the constant: (3x2 + 12x) + 4
Factor out the 3: 3(x2 + 4x) + 4
Complete the square: 3(x2 + 4x + 4) - 12 + 4 * (see below)
Factor: 3(x + 2)2 - 8
Pick out a, h, and K: a = 3 h = -2 k = -8
*Completing the square: divide the coefficient of the x-term (4x) by 2 and square the result:
4 / 2 = 2 ---> 22 = 4
place this number within the parentheses
because you have entered a 4 in the parentheses and the parentheses have a '3' before it,
you have added a '12', so you'll need to subtract a 12.
2) x - y2 + 8y = 13
Rearrange: - y2 + 8y = 13 - x
Multiply everything by -1: y2 - 8y = x - 13
Complete the squre: y2 - 8y + 16 = x - 13 + 16
Factor: (y - 4)2 = x + 3
Vertex: x = -3 y = 4
3) Like number 2.