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The equation ax2 + bx + c = 0, where a is a positive number, has no real solutions. Which of the following could be the vertex of the graph of f(x)

a) 0,0
b)-6,0
c)3,-2
d)0,4

1 Answer

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The vertex of a parabola given in the form of the equation y = ax^2 + bx + c is at the point (-b/2a, f(-b/2a)). In this case, since a is positive and the equation has no real solutions, the graph of f(x) must be a parabola that opens upwards and does not intersect the x-axis. Therefore, the vertex of the graph must be the lowest point on the y-axis.

Let's apply this to the answer choices:
a) 0,0 - This cannot be the vertex because it is on the y-axis, and we know that the parabola does not intersect the x-axis.
b) -6,0 - This cannot be the vertex because it is on the x-axis, and we know that the parabola does not intersect the x-axis.
c) 3,-2 - This could be the vertex because it is not on either axis, and it is the lowest point on the y-axis among the answer choices.
d) 0,4 - This cannot be the vertex because it is on the y-axis, and we know that the parabola does not intersect the x-axis.

Therefore, the answer is c) 3,-2.
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