Find Vertex and axis of symmetry for y = -(x - 5)^2 + 11

FMMurphy

Junior Member
Joined
Mar 12, 2006
Messages
51
I am confused by this problem because of all of the minuses. The problem is:

. . .y = -(x - 5)^2 + 11

I worked it by using FOIL:

. . .(x - 5)(x - 5)
. . .x^2 - 10x + 25

But because of the "minus" in front, I put:

. . .-x^2 + 10x - 25

Then I used the formula for the vertex, and put x = -b/(2a), which I did as:

. . .-(10) / (2(-1))
. . .-[(10) / (-2)]
. . .-(-5)

...or x = 5. Then I put x back into the expression:

. . .(-5)^2 + 10(5) - 25 + 11

This worked out would be y = 39, so the vertex is (5, 39).
 

camlax38

New member
Joined
Sep 25, 2006
Messages
11
Re: Find Vertex and axis of symmetry

FMMurphy said:
I am confused by this problem because of all of the minuses.
The problem is -(x-5)^2 +11

I workied it by using FOIL (x-5)(x-5) and I got x^2-10x+25 but with a minus in front I put -x^2 +10x -25 Correct

then I put x= -b/2a which I did as -(10/-2) which would be -(-5) or x=5
Then I put x into the original of (-5)^2 + 10(5) -25 +11
Worked out would be y =39
maybe looking at it like this will help you see it
-(b)/(2a) so it should be -(10)/(2*-1)=-10/-2=5

So replugging you get
-(5^2) + (10)(5) -25 +11 = -(25) +(50) -25+11=11 for your vertex
 

Mrspi

Senior Member
Joined
Dec 17, 2005
Messages
2,128
Re: Find Vertex and axis of symmetry

FMMurphy said:
I am confused by this problem because of all of the minuses.
The problem is -(x-5)^2 +11

I workied it by using FOIL (x-5)(x-5) and I got x^2-10x+25 but with a minus in front I put -x^2 +10x -25
then I put x= -b/2a which I did as -(10/-2) which would be -(-5) or x=5
Then I put x into the original of (-5)^2 + 10(5) -25 +11
Worked out would be y =39
I guess I am confused....

Is this your equation?

y = -(x - 5)<SUP>2</SUP> + 11

If so, it is in "vertex form", or y = a(x - h)<SUP>2</SUP> + k

The vertex of this parabola is at (h, k) and the axis of symmetry is x = h.

Please repost with more details if this is not what you meant.
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,943
FMMurphy said:
Then I put x back into the expression:

. . .(-5)^2 + 10(5) - 25 + 11
It should be:

. . . . .-(5)<sup>2</sup> + 10(5) - 25 + 11

The square is only on the 5, not the "minus" sign out front.

Note: The axis of symmetry is the vertical line through the vertex.

Eliz.
 
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