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17 November, 08:37

Suppose the equation ax^2+bx+c=0 has no real solution and a graph of the related function has a vertex that lies in the second quadrant.

Is the value of "a" positive or negative? Explain your reasoning.

Suppose the graph is translated so the vertex is in the fourth quadrant. Does the graph have any x-intercepts? Explain.

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  1. 17 November, 08:43
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    If it has no real solutions, that means the graph does not intersect the x axis

    since we have ax^2+bx+c=0, the parabola opens either up or down

    since the vertex is in the second quadrant (x is negative and y is positive in this reigon) and the graph does not cross the x axis, the parabola must open up

    if the value of 'a' is positive, then the parabola opens up

    so 'a' must be positive

    if it is translated to the 4th quadrant, then the vertex is now below the x axis

    it will now have 2 x intercepts because the vertex is in the 4th quadrant and look at a graph of a parabola opening up with vertex in 4th quadrant and seehow many time it crosses the x axis
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