Question 29169: could someone please help me?
solve by graphing:
x^2-4x+3=0
and -x^2+6x-5=0
Answer by Nate(3500)   (Show Source):
You can put this solution on YOUR website! Your Question:solve by graphing:
x^2-4x+3=0
and -x^2+6x-5=0
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We know that these are parabolas, so the graph will be "U" shaped!
Here is an easy way:
To find vertex it is (-b/2a,y). The vertex is the lowest or highest point in the parabola.
To find a and b and c, it is like this ax^2+bx+c=0. C just tells the y-intercept.
So, in x^2-4x+3=0, a=1, b=-4, and c=3.
vertext=(-b/2a,y)=((4/2),y)=(2,y)
To find y in the ordered pair for the vertex, you would plug in the x value in the ordered pair.
x^2-4x+3----original form
(2)^2-4(2)+3----plugging in the 2 for x from the vertex
(4)-8+3-----simplify
-1----answer
so the vertex is (2,-1)
Now we pick some ordered pairs. When the x value for an ordered pair is 3, we get (3,0). Other ordered pairs:(1,0)(4,3)(0,3)
Now, graph those points and connect to get a "U" (parabola.)
Since the equation is equal to zero, the answers are the points that touch the x-axis on the parabola. The answers are 1 and 3!
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I will do the same for the other equation, but shorter.
-x^2+6x-5=0
Vertex(3,4)
Other points:(2,3)(4,3)(1,0)(5,0)
Since the equation is equal to zero, the answers are the points that touch the x-axis on the parabola. The answers are 1 and 5!
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