help pleasede

[wraith drum]

For Q2 complete the square so you can find the turning point and also the x-intercepts

[fierce yoke]

How do I complete the square🫣

[wraith drum]

2x^2 - 8x - 5
2(x^2 - 4x) - 5
Complete the square inside the brackets
Half the co efficient of x and then square it on the outside of the brackets
2[(x - 2)^2 - 2^2] - 5
2[(x-2)^2 - 4] - 5
Multiply everything inside [] by the 2
2(x-2)^2 - 8 - 5
2(x-2)^2 - 13
For the turning point co ordinates it needs to be in the form
a(x - h) + k
So we flip the sign of the 2 and take the -13
(2, -13)
Now for the x-intercepts we set the equation = 0
2(x-2)^2 - 13 = 0
2(x-2)^2 = 13
(x-2)^2 = 13/2
x - 2 = +-root(13/2)
x = 2 + root(13/2)
x = 2 - root(13/2)

[fierce yoke]

Your amazing

Thank you

[fierce yoke]

Hello again

Im kinda stuck here

[wraith drum]

Take half of the coefficient of x
So it’s
-3[(x - 3)^2
And then we want to copy that over and square it so if we was to expand it it would = x^2 - 6x
(x - 3)^2 - 3^2
(x-3)^2 - 9
(x^2 - 6x + 9) - 9 = x^2 - 6x
As shown^
So we have
-3[(x-3)^2 - 9] + 9

Then you’d just multiply everything in the square brackets by -3

[fierce yoke]

Thank youu

Would the turning point be -3,0 then?

[fierce yoke]

Or was my working out wrong

[wraith drum]

Idk I haven’t actually tried working it out

[fierce yoke]

Eh ill just go with -3,0 this stuff is too confusing

Do you understand this one too?

[wraith drum]

Yea it’s just asking you to complete the square until it looks like that

[fierce yoke]

Ughh

[fierce yoke]

Is this the correct way of doing it

[wraith drum]

Is that b = 8?

[fierce yoke]

No b=0

[wraith drum]

Well if you complete the square you get
(x - 3)^2 - 9 + 1
(x-3)^2 - 8
a = 3
b = 8