Help with a simple question (1 Viewer)

[abdog]

Find integers a and b such that x^2 + 6x + 14 = (x+a)^2 + b.

 

[lochnessmonsta]

Expand RHS ---> x^2 + 2ax + a^2 + b

Equating coefficients of x ---> 2a = 6
a = 3

Equating constants ---> a^2 + b = 14
9 + b = 14
b = 5

Therefore a=3, b=5

 

[Peeik]

x^2+6x+14
= x^2+6x+9+5
= (x^2+6x+9)+5
= (x+3)^2 +5

By comparing coeff and constants:

a=3 and b=5

 

[abdog]

How do you go from 14 to 9+5? Do you just guess it?

Also, I have another question. Consider the parabola y=x^2-4x+8. Find the coordinates of the focus.

 

[iBibah]

How do you go from 14 to 9+5? Do you just guess it?

In order to complete the square, you need to divide 6 by 2, then square it, which gives 9. Therefore you split 15 into 9+5 in order to complete the square.

 

[abdog]

Ah, I see. Can you answer my other question?

 

[iBibah]

Ah, I see. Can you answer my other question?

Put it in the form x^2=4ay.

 

[abdog]

And how exactly do you do that again?

 

[Kurosaki]

How do you go from 14 to 9+5? Do you just guess it?

Also, I have another question. Consider the parabola y=x^2-4x+8. Find the coordinates of the focus.

x^2-4x=y-8
x^2-4x+4=y-4
(x-2)^2=4*0.25(y-4)
So we see that the parabola is an upwards pointing one- with a focal length of 0.25 units.
since it is in the form (x-h)^2=4a(y-k), the focus will be up 0.25 units from the vertex. so the focus is, like (2,4.25) ?

 

[abdog]

How do you get a? The a in 4a.