Completing The Square (solving, finding vertex, etc)

[wiishesssss]

11. Complete the square for this expression:

. . .4s^2 + s + 2

b = 1, so (1/2) ^ 2 = 1/4 = 0.25

Therefore, 4s^2 + 1s + 0.25 is a perfect trinomial

I need help from there after for that problem.

17. Complete the square to find the vertex of the parabola:

. . .y = x^2 + 6x + 3

(b/2)^2 = (6/2)^2 = (3)^2 = 9

x^2 + 6x + 9 is a perfect square trinomial

x^2 + 6x + 9 -9 + 3

Therefore you have, (x+3) ^ 2 - 6

From there now how do you determine what is the vertex in the problem?

37. Solve by using the quadratic formula:

. . .n ^ 2 - 4n - 12 = 0

a = 1, b = -4, c = -12

n = (-b +/- Sq. Rt. [b^2 - 4ac]) / (2a)

n = (4 +/- Sq. Rt. [(-4)^2 - 4(1)(-12)]) / [2(1)]

n = (4 +/- Sq. Rt. [16 + 48]) / 2

n = (4 +/- Sq. Rt. [64]) / 2 . . .=>Sq. Rt. of 64 = 8

n = (4 +/- 8) / 2

n = (4 + 8) / 2 = 12/2 = 6

n = (4 - 8) / 2 = -4/2 = -2

n = 6, -2

I am pretty sure that was done correctly, but if I can get a confirmation, I would appreciate it. Thank you

 

[stapel]

11. Complete the square for... 4s^2 + s + 2

4s^2 + 1s + 0.25 is a perfect trinomial...

Not actually: 4s² = (2s)², and 1/4 = (1/2)². But for the above to be a perfect square, you need the middle term to be 2(2s)(1/2) = 2s, not 1s.

Don't forget to factor out whatever is multiplied on the squared term. That is always the first step (and is generally what makes things messy):

. . . . .4s² + s + 2

. . . . .4(s² + s/4) + 2

. . . . .(1/2)(+1/4) = +1/8 (keep track of that sign)

. . . . .(+1/8)² = 1/64

Now add inside and subtract outside, accounting for that "4" in front:

. . . . .4(s² + s/4 + 1/64) + 2 - 4(1/64)

. . . . .4(s + 1/8)² + 2 - 1/16

. . . . .4(s + 1/8)² + 31/16

17. Complete the square to find the vertex of the parabola:

. . .y = x^2 + 6x + 3...

Therefore you have, (x+3) ^ 2 - 6

From there now how do you determine what is the vertex in the problem?

The vertex form is "y = a(x - h)² + k", where the vertex is the point (h, k). So just read the vertex off the equation.

37. Solve by using the quadratic formula: n ^ 2 - 4n - 12 = 0....

n = 6, -2

Easy check, since this factors:

. . . . .n² - 4n - 12 = (n - 6)(n + 2) = 0

. . . . .n - 6 = 0 or n + 2 = 0

. . . . .n = 6 or n = -2.

The solution checks.

Eliz.

 

[wiishesssss]

So then for 17, the vertex is (-3,-6)

y = a (x-h) ^ 2 + k
(x-3) ^ 2 - 6


so for h ...

-(3) = -3



and for k ..

+(-6) = -6



is that right?

 

[stapel]

So then for 17, the vertex is (-3,-6) [from]y = 1(x - 3) ^ 2 - 6

Since the vertex form is "y = a(x - h)² + k", and since x - (-3) = x + 3, then the x-coordinate for the vertex cannot be -3; it should be +3: (h, k) = (3, -6)

Eliz.