× Didn't find what you were looking for? Ask a question
Top Posters
Since Sunday
6
y
2
m
2
m
2
u
2
m
2
B
2
M
2
e
2
k
2
N
2
y
2
Home Q & A Board Science-Related Homework Help Mathematics
New Topic  
Dynamicb Dynamicb
wrote...
Posts: 1
Rep: 0 0
12 years ago
Can someone tell me the answer to this question and explain why or why not all equations of parabolas can be expressed in factored form?
Read 589 times
4 Replies

Related Topics

Replies
wrote...
#1
12 years ago
some equations are not factorable. in this case you have to leave in its

y=ax^2+bx+c form
wrote...
#2
12 years ago
Yes, they can be.
If they are of the form y=ax^2+bx+c, by completing the square, they can be expressed as y=d(x-h)^2+k, where (h,k) is the vertex.
Conversely, if they are of form y=d(x-h)^2+k, they can be expressed as
y=ax^2+bx+c by expanding the vertex form to the quadratic form.
wrote...
#3
12 years ago
All equations of parabolas cannot be expressed in factored form.  Sometimes, the roots of quadratic equations are irrational or imaginary.  Therefore, it will not be possible for the equation to be factored so it will not be possible to express it in factored form.
wrote...
#4
12 years ago
YES! but NOT in a binomial form like, (x-1)(x+1)..

we just did parabolas as our last unit in precal and the teacher NEVER showed us a parabola in a binomial factored form just like the one above.  Is was usually/always in y=a(x-h)² +k  where "a" cannot equal to zero and (h,k) is the vertex

i think if it was in a binomial factored form, you're looking for zeros
New Topic      
Quick Reply
Explore
Post your homework questions and get free online help from our incredible volunteers
  507 People Browsing
Start New Topic Take the Tour CoursesNew Study Tools Topics Trending Browse by Textbook
Live Chat
Related Images
  
 1868
  
 348
  
 1278
Your Opinion
How often do you eat-out per week?
Votes: 81
Previous poll results: Who will win the 2024 president election?