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

NIbberlat Jun 30, 2020

#1**0 **

1) Rewrite 3x^{2} + 12x + 4 into the form a(x - h)2 + k

3x^{2} + 12x + 4

Separate the x-terms from the constant: (3x^{2} + 12x) + 4

Factor out the 3: 3(x^{2} + 4x) + 4

Complete the square: 3(x^{2} + 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 ---> 2^{2} = 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 - y^{2} + 8y = 13

Rearrange: - y^{2} + 8y = 13 - x

Multiply everything by -1: y^{2} - 8y = x - 13

Complete the squre: y^{2} - 8y + 16 = x - 13 + 16

Factor: (y - 4)^{2} = x + 3

Vertex: x = -3 y = 4

3) Like number 2.

geno3141 Jun 30, 2020