Question 287955: I need your help on Solving Quadratic Equations by Graphing. I get the graphing part but it is just the equation y^2-4y+1.
I do know how to get the vertex and then I get blank after the I get vertex.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! I need your help on Solving Quadratic Equations by Graphing. I get the graphing part but it is just the equation y^2-4y+1.
I do know how to get the vertex and then I get blank after the I get vertex.
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Equations are solved, and all equations have equal signs.
y^2-4y+1 is not an equation, it's a polynomial.
If it were
y^2-4y+1 = 0, or = f(y) it could be solved for y.
I'll assume that's what you meant, tho it's not my job to try to guess what you and others meant to say. It's your job to say what you're trying to do.
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y^2-4y+1 = 0
The vertex is on the line of symmetry, and that's -b/2a
= 4/2 --> y = 2
f(2) = 4 - 8 + 1 = -3
The vertex, on an x-y plane, is (-3,2)
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I don't know what you mean by "and then I get blank after the I get vertex."
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The online solver uses x, sub y for x
| Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc) |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=12 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: 3.73205080756888, 0.267949192431123.
Here's your graph:
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y = 2+sqrt(3)
y = 2-sqrt(3)
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