Finding p and q when a parabola crosses the x-axis at -0.5 and 2

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Peter G.
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Hi,

The diagram represents the graph of the function: f(x) (x-p) (x-q)

The diagram shows a quadratic graph that intersects the x-axis twice: at -0.5 and 2.

It then asks us for the value of p and q.

I know the answers are: 0.5 and -2

But I am confused on how to determine which is which :mad:

Thanks,
Peter G.
 
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Of course, there is no way to tell which is which !
 
Yeah, that's what I thought but the book put specific answers for p and q... Not very good question I guess :-p
 
Oh, and one thing:

The diagram shows part of the graph of the function:
ax2+bx + c

The graph is a upside down parabola fully to the right of the y axis, and its vertex only touches the x axis. It then asks:

The value of a, c, and b - Whether they are positive or negative.

Value of a: Negative because the curve is upside down
Value of c: Negative because it cuts the y-axis below the x axis
Value of b: I am confused with this: I tried playing around completing the square several times and as I increased the value of b, I increased how much it moved to the left or right. Furthermore, if my value of b was positive, within the perfect square, the value was also positive, so I concluded. If the graph moved to the right, my value within the brackets should be negative but the book says positive, could you explain?

Thanks
 
The x-coordinate of the vertex is -b/(2a).
 
So b must be positive the vertex is in a location (to the right of the y axis) where the x values are positive? b must be positive because we have a as negative and the result of the multiplication must be positive?