Question

For the following parabola equation, find the x-intercepts and the coordinates of the
vertex.
y= -x^2 -2x + 3

Answer 1

are you familiar with completing the square?

Answer 2

to find the x-intercept, simply let y=0.

Answer 3

No I dont know completing the squares I've seen it but I could never get it.

Answer 4

Okay, then are you familiar with differentiation?

Answer 5

like... factoring?

Answer 6

um, dy/dx. If not, then I guess you'll have to start from completing the square.

Answer 7

ya lets start from completing the square idk what dy/dx is.

Answer 8

\((x+1)^2 = x^2 +2x +1\)
\((x+1)^2+3 = x^2 +2x +4\)
So, completing the square is simply reverting the right side back to the left side.

Basically , we try to guess what number fits the square
\(y= -x^2 -2x + 3 \)

now, take out the neg so it's easier.
\(y= -(x^2 +2x - 3) \)

Notice that x^2 +2x just almosts fits a perfect square.
so, we add 1 to it. but to compensate, we minus 1 too, so it isn't like we're pulling it out of thin air.
\(y= -(x^2 +2x+1-1 - 3) \)
\(y= -(x^2 +2x+1 - 4) \)

Now, \(x^2 +2x+1=(x+1)^2\)
\(y= -( (x+1)^2- 4) \)
\(y= - (x+1)^2+4 \)

After completing the square, you can see that the max value for y will be when \((x+1)^2=0\)
so, x=-1, y=4, vertex is (-1,4)

you follow?

Answer 9

woah, lets see

Answer 10

ya I dont get how you get from y=−(x2+2x+1−4) to
y=−((x+1)^2 −4) because with (x^2 + 2x +1 -4) you should get x^2 +2x -3

Answer 11

yes. but in completing the square,
we want to make an quad expression into a perfect square, with some thing behind.
(x^2 + 2x +1 -4) looks very much like ((x+1)^2 −4)

Answer 12

I dont think im getting it and my mind is getting flustered, I need a break.. Thank you for trying anyway.

Answer 13

lol okaay