How to sketch this parabola problem..

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candy101

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Joined
Oct 4, 2009
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43
How to sketch this problem...
how would i sketch this==> y=x^2+6x+5


i don't if this is correct but this is how i am trying to do it===> i label the numbers A,B, C
A=1
B=6
c=5

then i did -b/2a===> -6/2(1)= -3

then i just plug that value in for x==>and i got -22
so my vertex is (-3,-22) ==> is this correct?
 
candy101 said:
How to sketch this problem...
how would i sketch this==> y=x^2+6x+5


i don't if this is correct but this is how i am trying to do it===> i label the numbers A,B, C
A=1
B=6
c=5

then i did -b/2a===> -6/2(1)= -3

then i just plug that value in for x==>and i got -22 <<< This is not correct
so my vertex is (-3,-22) ==> is this correct?

Right idea...
Click to expand...
 
awesome! (i always manage to mess up some where along the line)

umm...i think the x-intercept is (-1,0) and (-5,0)

and the y- is (0,5)

and i got a question (reading up on these kind of questions it says find the x- intercepts if there is any)

how would i know if there is any x-intercepts? and if they aren't?



Thank you,
candy
 
candy101 said:
awesome! (i always manage to mess up some where along the line)

umm...i think the x-intercept is (-1,0) and (-5,0)

and the y- is (0,5)

and i got a question (reading up on these kind of questions it says find the x- intercepts if there is any)

how would i know if there is any x-intercepts? and if they aren't?



Thank you,
candy
Click to expand...

How did you find x-intercepts?
 
i got the x-intercepts by setting y=o
and i factor x^2+6x+5
which gave me x+1 and x+5
then i set it equal to 0 so i got x=-1 and x=-5 ==> is it not right?
 
candy101 said:
I got the x-intercepts by setting y=o
and I factor x^2+6x+5
which gave me x+1 and x+5
then I set it equal to 0 so I got x=-1 and x=-5 ==> is it not right?
Click to expand...
It's correct , Nevada ...
 
You do not have x-intercept when

\(\displaystyle B^2 - 4\cdot A \cdot C\: < \: 0\)
 
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