Question

y=-x^2 +2x -5 graph and label the vertex and any intercepts that exist

Answer 1

Have you considered completing the square?

Answer 2

no I tried the quadtraic formula but I think I did it wrong because I feel like u have to do something special with the negative, like take it out.
so I factor out - and get y=-1 (x^2 + 2x -5) and using the quadtratic formula I get \[2\pm \sqrt{-16}\over 2\]

Answer 3

so now im like -16? that would leave me with no real solution no? and just move on to getting the vertex which I got to be 1,-2. So I have a vertex but no x intercepts?

Answer 4

Anyone? here?

Answer 5

The Quadratic Formula will show youthe x-values associated with y = 0, if there are any. This does solve part of the problem, but certainly not all of it. Can you "Complete the Square" on this one?

Answer 6

no, I dont know how to but I tried. So going from the original y= -x^2 + 2x - 5 I tried by taking 1/2 of b from this -x^2 + 2x -5 and I got 1, and so I square that and I get -x^2 +2x +1 = y +1 and I rewrite is to its -(x-1)^2 = y +1and I guess you could put the 1 back on the left side so its -(x-1) (x-1) +1 = y so...
what? x= 1, x= 1 and x=-1?

Answer 7

?? Let's be more systematic about it.

\(y = -x^{2} + 2x - 5\)

It generally makes more sense if the leading coefficient is oen (1).

\(y = -(x^{2} - 2x + 5)\)

That "-2" is our hint. Half of that, squared is what we need, not that "+5"

-2/2 = -1
(-1)^2 = 1

\(y = -(x^{2} - 2x + 1 - 1+ 5)\)

Just adding and subtracting the same thing.

Simplify

\(y = -(x^{2} - 2x + 1 + 4)\)

Split out the piece we don't need. (Distributive Property)

\(y = -(x^{2} - 2x + 1) - 4\)

Rewrite the way we need it.

\(y = -(x - 1)^{2} - 4\)

In this form, the Vertex is rather staring at us.

Answer 8

ooo lol I was typing out this whole thing and rereading what you put and I finally understood ur way. But for this vertex and it staring at us I still dont get that part if vertex= -b/2a, b= 2 and a =-1, so -2/2(-1) so vertex equal (1,y) so far and then plug in
-x^2 + 2x - 5 so -1^2 + 2(1) -5 -> 1 + 2 -5 = -2 so vertex is 1, -2 ?

Answer 9

-b/2a is the x-coordinate of the vertex. You must substitute this value of x to find the associated y-coordinate. You have the idea, but something went wrong along the way. (1,-4)

Completing the square puts it in Vertex Form where some informatin concerning its properties are more apparent. (1,-4) is quite obvious in the Vertex Form.

Answer 10

ya idk man I did this over and looked at other examples I'm not seeing what you see. Also how come I couldnt do this in quadratic formula?

Answer 11

thank you tkunny but uve seen to abandon me a long time ago, and ur trying to teach me on another lvl I appreciate it but its not really helping me by throwing me off a cliff and expecting me to fly. Anyone else can Help me?

Answer 12

Actually, Server Problems. I keep getting disconnected. Quite annoying.

Completing the Square is not another level. You need to gain this skill.
Read again, The answer to your Quadratic Formula question is already inteh discussion.