SOLUTION: Given a>0 for what values of k does y=a(x-k)^2+k have? a) two x-intercepts ? b) one x intercept ? c) no x-intercepts? Please explain why for each one. Thank you.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Given a>0 for what values of k does y=a(x-k)^2+k have? a) two x-intercepts ? b) one x intercept ? c) no x-intercepts? Please explain why for each one. Thank you.       Log On


   

Question 1003127: Given a>0 for what values of k does y=a(x-k)^2+k have?
a) two x-intercepts ?
b) one x intercept ?
c) no x-intercepts?
Please explain why for each one. Thank you.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The x-intercepts, if any, are the points where y=0 (or the values of x for those points).

IF YOU WERE NOT TAUGHT ABOUT THE DISCRIMINANT OF A QUADRATIC EQUATION:
The x-intercepts, if any, are the points where y=0 (or the values of x for those points).
For k=0 the function is y=ax%5E2 , and has one x-intercept at x=0
For k%3E0 , since %28x-k%29%5E2%3E=0 for all values of x ,
y=a%28x-k%29%5E2%2Bk%3E=0%2Bk=k%3E0 , meaning that there is no x-intercept.
For k%3C0 , there are two intercepts, because -k%2Fa%3E0 , so sqrt%28-k%2Fa%29%3E0 , and
y=0-->a%28x-k%29%5E2%2Bk=0-->a%28x-k%29%5E2=-k-->%28x-k%29%5E2=-k%2Fa-->system%28x-k=sqrt%28-k%2Fa%29%2C+%22or%22%2Cx-k=sqrt%28-k%2Fa%29%29-->x=k+%2B-+sqrt%28-k%2Fa%29 .

IF YOU WERE TAUGHT ABOUT THE DISCRIMINANT OF A QUADRATIC EQUATION:
The reasoning above is still valid, but you may be expected to talk about the discriminant in an answer.
That makes it complicated, but here it goes:
y=a%28x-k%29%5E2%2Bk<--->y=a%28x%5E2-2kx%2Bk%5E2%29%2Bk<--->y=ax%5E2-2akx%2Bak%5E2%2Bk<--->y=ax%5E2-2a%5E2kx%2B%28ak%5E2%2Bk%29
For a quadratic equation of the form ax%5E2%2Bbx%2Bc=0 ,
discriminant=b%5E2-4ac .
If discriminant=b%5E2-4ac%3E0 , ax%5E2%2Bbx%2Bc=0 has two solutions;
if discriminant=b%5E2-4ac=0 , ax%5E2%2Bbx%2Bc=0 has one solution, and
if discriminant=b%5E2-4ac%3C0 , ax%5E2%2Bbx%2Bc=0 has no solution.
Applied to y=ax%5E2-2a%5E2k%2B%28ak%5E2%2Bk%29=0 ,
where b=-2ak and c=ak%5E2%2Bk ,
discriminant+=%28-2ak%29%5E2-4a%28ak%5E2%2Bk%29=4a%5E2k%5E2-4a%5E2k%5E2-k=-k , so
if -k%3E0<-->k%3C0 , y=0 has two solutions, and y=a%28x-k%29%5E2%2Bk has two x-intercepts;
if -k=0<-->k=0 , y=0 has one solutions, and y=a%28x-k%29%5E2%2Bk has one x-intercepts, and
if -k%3C0<-->k%3E0 , y=0 has no solutions, and y=a%28x-k%29%5E2%2Bk has no x-intercepts.