SOLUTION: This is a parabola question. I really need help solving. if you could please show me the work and also tell me how to solve. thank you! :) without drawing the graph of the given

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: This is a parabola question. I really need help solving. if you could please show me the work and also tell me how to solve. thank you! :) without drawing the graph of the given      Log On


   

Question 256575: This is a parabola question. I really need help solving. if you could please show me the work and also tell me how to solve. thank you! :)
without drawing the graph of the given equation determine
A) how many x-intercepts the parabola has
B) whether its vertex lies above or below or on the x-axis.
show your work.
1) y=x(squared) - 5x +6
2) y= -x(squared) + 2x -1
Found 2 solutions by richwmiller, drk:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
It would help if you put more time into copying the question correctly and less time crying.
Notice how squared is correctly written. Notice how 1) appears by enclosing the problem in three "{" and three "}" curly braces. Next to the p key
1)+y=x%5E2-5x%2B6
2) y=-x^2+2x-1
The rules of the website say
One problem per submission.
No similar problems and
Limit of 4 submissions daily
You broke two of the rules already.
One and two are similar and are two problems.
Whereas A and B can be considered one problem
Here is the solution to 1)
The determinant tells how many x roots there are.
See description below
vertex lowest point (5/2, -1/4) slightly below the x axis
the x of the vertex is -b/2a find that and plug it in to find y
-(-)5/2(1)=5/2
plug 5/2 for x and get -1/4 for y in
y=x^2-5x+6
y=5/2^2-5*5/2+6
y=25/4-25/2+24/4
y=(25-50+24)/4
y=-1/4
BTW y=x^2-5x+6 can be factored into y=(x-3)(x-2)
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-5x%2B6+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-5%29%5E2-4%2A1%2A6=1.

Discriminant d=1 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+1+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+1+%29%29%2F2%5C1+=+3
x%5B2%5D+=+%28-%28-5%29-sqrt%28+1+%29%29%2F2%5C1+=+2

Quadratic expression 1x%5E2%2B-5x%2B6 can be factored:
1x%5E2%2B-5x%2B6+=+1%28x-3%29%2A%28x-2%29
Again, the answer is: 3, 2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-5%2Ax%2B6+%29

Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
without drawing the graph of the given equation determine
A) how many x-intercepts the parabola has
ANS: There are three options (i) 0 intercepts; (ii) 1 intercept. (iii) 2 intercepts. The way to determine this is by the discriminant. It is
D+=+sqrt%28b%5E2+-+4ac%29
If D < 0, then 0 intercepts
If D = 0, then 1 intercept
If D > 1, then 2 intercepts
--
B) whether its vertex lies above or below or on the x-axis.
show your work.
1) y=x^2 - 5x +6
D+=+sqrt%28b%5E2+-+4ac%29
D+=+sqrt%2825-4%2A1%2A6%29
D+=+sqrt%281%29
D > 0, so 2 intercepts
--
2) y= -x^2 + 2x -1
D+=+sqrt%28b%5E2+-+4ac%29
D+=+sqrt%284-4%2A-1%2A-1%29
D+=+sqrt%280%29
D = 0, so 1 intercept