SOLUTION: Please help me! Cannot figure this out. it says: without drawing the graph of the given equation,determine a)how many x-intercepts the parabola has b)whether its vertex lies above

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Question 131281This question is from textbook structure and method
: Please help me! Cannot figure this out. it says: 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.
y=x^2-5x+6
Please and Thank You=) This question is from textbook structure and method

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
I'll answer part b first:

The vertex of a parabola which equation is in the form y=f%28x%29=ax%5E2%2Bbx%2Bc is at the point (-b%2F%282a%29,f%28-b%2F%282a%29%29). For your equation, a = 1, b = -5, and c = 6.

%28-%28-5%29%29%2F2%281%29=5%2F2



So the vertex is located at: (5%2F2,-1%2F4). The y-coordinate, -1%2F4 is less than zero, therefore the vertex is below the x-axis.

Part a:
Two ways to answer this: Since the lead coefficient is greater than zero, the parabola must be concave up (makes a valley instead of a hill). A concave up parabola with a vertex below the x-axis must cross the axis in two places. If the y-coordinate of the vertex were greater than zero, the parabola wouldn't cross the x axis at all, and if the y-coordinate is equal to zero, then the parabola intersects the x axis in exactly 1 point -- the axis is then said to be tangent to the parabola.

The other way to answer the question is to set the polynomial equal to zero and solve for x. If you get two real number roots, then the graph must have two x-intercepts.

x%5E2-5x%2B6=0

%28x-2%29%28x-3%29=0

x=2 or x=3, and the graph does indeed have two x-intercepts, one at (2,0) and the other at (3,0).


Did you notice that the x-coordinate of the vertex is exactly half-way between the two intercepts?

Had you obtained 2 real and equal roots, which is the result you get when the trinomial is a perfect square, then the axis would be tangent at the vertex and you would have a single point of intersection. Had you obtained a complex conjugate pair as a solution set to the equation, then you would have no points of intersection with the x-axis.

It really helps to see this visually, so I'm going to give you the graph of your equation along with two others, y=x%5E2-5x%2B%2825%2F4%29 (a perfect square that has the x-axis as a tangent), and y=x%5E2-5x%2B7 (no real roots and will not intersect the x-axis)