Question 227719: 1)If the graph of y=x2-6x+c is tangent to the x-axis, then c equals
(1) 3
(2) -3
(3) 9
(4) -9
6)Which parabola intersects the x-axis in two distinct points?
(1) y=(x+5)^2
(2) y=(x-5)^2
(3) y=x^2-25
(4) y=x^2+25
9)If the graph of y=4x2+bx=1 intersects the x-axis in two distinct points,then b may be equal to
(1) 3
(2) 4
(3) -4
(4) 5
These are the problems i cant solve. Please show work and i dont know if this is the right topic but i know it Algebra 2 thanks.
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! 1)If the graph of y=x2-6x+c is tangent to the x-axis, then c equals...
The only way for a parabola to be tangent to the x-axis is if the vertex is on the x-axis. And the x-coordinate of the vertex of a parabola is -b/2a, which in this case is -(-6)/2(1) = 3. And on the x-axis the y-coordinate is 0. So all we have to do is substitute 3 for x and 0 for y and solve for c:
6)Which parabola intersects the x-axis in two distinct points?
(1) y=(x+5)^2
(2) y=(x-5)^2
(3) y=x^2-25
(4) y=x^2+25
Unlike #1, we do not want the vertex on the x-axis. Since all of these equations have an "a" (coefficient of  ) which is positive, these are all parabolas which open upward. So in order to intersect the x-axis twice, we need a vertex which is below the x-axis. So find the vertex of each. (1) and (2), you will find, have a vertex on the x-axis (like problem #1). And (4) has a vertex above the x-axis. This parabola does not intersect the x-axis at all. Only (3) have a vertex below the x-axis.
9)If the graph of y=4x2+bx=1 intersects the x-axis in two distinct points,then b may be equal to
(1) 3
(2) 4
(3) -4
(4) 5
Like problem #6 we need a vertex below the x-axis. And in order for the vertex to be below the x-axis, the y-coordinate of the vertex has to be negative. So what we can do is try each possible b. Then find the x-coordinate of the vertex, -b/2a and then see if the y-coordinate for that x in negative. If we try (1), b=3:
x-coordinate of the vertex: -3/2(4) = -3/8
y-coordinate of the vertex = 
Since the y is positive, this is not our solution.
Repeating this process for the other possible b's we should find that only (4), b=5, results in a y-coodinate which is negative. So only with b=5 do we get a vertex below the x-axis and, therefore, a parabola that intersects the x-axis twice.
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