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.

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

For the first question the answer is a, for the second c, for the third d, for the 4th b, for the 5th a