Quadratics

[kris.yarmak123]

A quadratic solution y=(x-p)^2 passes through the points (-5,24) and (-9,0). The value of p must be ____
Plz help I forgot how to solve it

 

[MarkFL]

Are you sure the problem is copied correctly?

 

[kris.yarmak123]

Oh it’s supposed to be y=(x-p)^2+q . Damn sorry

 

[Deleted member 4993]

Oh it’s supposed to be y=(x-p)^2+q . Damn sorry

"A quadratic solution y=(x-p)^2 passes through the points (-5,24) and (-9,0). The value of p must be ____ "

use x= -5 and y = 24 in the given equation. What do you get?

 

[kris.yarmak123]

I get 24=(5+p)^2+q

 

[kris.yarmak123]

Or y=(x+5)^2+24

 

[Deleted member 4993]

Or y=(x+5)^2+24 .....................No - none of those!!

we have x = -5 and y = 24.

using y=(x-p)^2+q, Replace x with -5 and y with 24 → and we get,

24 = (-5 - p)^2 + q

p^2 + 10*p + q = -1 ............................................(1)

Now again, we have x = -9 and y = 0.

using y=(x-p)^2+q, we get = ????

 

[kris.yarmak123]

we have x = -5 and y = 24.

using y=(x-p)^2+q, Replace x with -5 and y with 24 → and we get,

24 = (-5 - p)^2 + q

p^2 + 10*p + q = -1 ............................................(1)

Now again, we have x = -9 and y = 0.

using y=(x-p)^2+q, we get = ????

0=(-9-p)^2+q

 

[kris.yarmak123]

Then -81=p²+18p+q

 

[kris.yarmak123]

Ohhh got it, p=-10

 

[kris.yarmak123]

Thank you so much!

 

[Deleted member 4993]

Thank you so much!

and q = ?

Did you check your answer?

 

[kris.yarmak123]

For q I did 24=(-5-(-10))^2+q
24=25+q
q=-1. I substituted the numbers in the first equation and found out that 24=24 therefore it's correct. I also checked the answer sheet which says that p=-10