Question 198552: Look at the graph above and comment on the sign of D or the discriminant. Form the quadratic equation based on the information provided and find its solution.
I am not sure how to upload the graph, but the numbers posted are (1,0) (5,0) (3-4). Any help would be greatly appreciated.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Look at your graph. If the entire graph is on one side or the other of the x-axis, that is, the graph does not intersect the x-axis anywhere, then the discriminant is less than zero,  . If the graph intersects the x-axis in two points, then the discriminant is greater than zero,  . If the graph is tangent to the horizontal axis, which is to say that there is only one intersection point and that point is the vertex of the parabola, then the discriminant is zero,  .
This all makes good intuitive sense because, in order for there to be real number zeros of the function, the graph has to intersect the x-axis. So, no intersection, no real roots. One intersection means a perfect square trinomial with one real root with a multiplicity of two. Two intersections, two distinct real roots. So there is a direct correlation between the zeros of a quadratic equation and the graph of the corresponding quadratic function. Don't you just love it when a plan comes together?
With the given points, there are two points of intersection with the x-axis, namely (1, 0) and (5, 0). So:
and
Giving us 2 factors for a potential quadratic trinomial. Let's test.
(x - 5) = x^2 - 6x + 5)
The question now is if (3, -4) is a point on the graph of the derived quadratic.
^2 - 6(3) + 5 =^? -4)
The reason that we had to check to see if (3, -4) was on the graph is that there are an infinite number of quadratic trinomials that would have the given roots, 1 and 5, and the set of trinomials could be described by:
)
where  is any non-zero real number. But there is only one value for that would create a trinomial with a graph that also contains the point (3, -4). In this case, we lucked out because 
Compare this to the situation where the third point is (3, 4). Here we would have
^2 - 6(3) + 5) = k(-4) \ \ \Rightarrow\ \ k = -1)
and the quadratic function would then have been:
John
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