Rondragon123 Lol are you a kid? Just multipal each number on the first ( ) by each number on the second ( ) Reply February 18, 2013
FrostyGun First expand the equation (x+1)(x-5) = x^2 -4x -5 Now use the Formula -b/2a Where b = -4where a = x^2 = 1 -(-4)/2(1) = 2 Put the 2 into the equation of x^2 -4x -5 (2)^2 -4(2) -5 = 9 therefore your vertex is 2, -9 Reply February 18, 2013 - edited
goobirdx use derivatives x^2-4x-5 then 2x=4 x=2 now plug it in the original and you'll get -9 (2,-9) Reply February 18, 2013 - edited
NonSonoFronz [quote=megatron1121]What I can remember. X intercepts are -1 and 5, axis of symmetry in 2 but I can't figure out the -9 part[/quote] If the axis of symmetry is 2, you can just plug it in to your function to find y. Reply February 18, 2013 - edited
megatron1121 What I can remember. X intercepts are -1 and 5, axis of symmetry in 2 but I can't figure out the -9 part Reply February 18, 2013 - edited
daStrike [quote=ZOMGitjon]set both to 0x+1=0x-5=0[/quote] isn't that intercept Reply February 18, 2013 - edited
MatthewDough [url=http://www.uiowa.edu/~examserv/mathmatters/tutorial_quiz/geometry/findingvertexofparabola.html]I advise you read this first.[/url] [url=http://www.wolframalpha.com/input/?i=%28x%2B1%29%28x-5%29]The vertex is at (2,-9).[/url] Reply February 18, 2013 - edited
Guardians Vertex should be at -1, -5 if you want to double check your answer. Reply February 18, 2013 - edited
9 Comments • Newest first
Lol are you a kid? Just multipal each number on the first ( ) by each number on the second ( )
First expand the equation
(x+1)(x-5) = x^2 -4x -5
Now use the Formula -b/2a
Where b = -4
where a = x^2 = 1
-(-4)/2(1) = 2
Put the 2 into the equation of x^2 -4x -5
(2)^2 -4(2) -5 = 9
therefore your vertex is 2, -9
use derivatives x^2-4x-5 then 2x=4 x=2 now plug it in the original and you'll get -9 (2,-9)
[quote=megatron1121]What I can remember. X intercepts are -1 and 5, axis of symmetry in 2 but I can't figure out the -9 part[/quote]
If the axis of symmetry is 2, you can just plug it in to your function to find y.
What I can remember. X intercepts are -1 and 5, axis of symmetry in 2 but I can't figure out the -9 part
[quote=ZOMGitjon]set both to 0
x+1=0
x-5=0[/quote]
isn't that intercept
[url=http://www.uiowa.edu/~examserv/mathmatters/tutorial_quiz/geometry/findingvertexofparabola.html]I advise you read this first.[/url]
[url=http://www.wolframalpha.com/input/?i=%28x%2B1%29%28x-5%29]The vertex is at (2,-9).[/url]
Vertex should be at -1, -5 if you want to double check your answer.
set both to 0
x+1=0
x-5=0