If 7−px−x2=16−(q+x)2, what are p and q?
Question
Given that y=7−px−x2=16−(q+x)2 for all values of x, where p and q are both positive,
- calculate the values of p and q,
- state the maximum value of y and the value of x at which it occurs,
- find the range of values of x for which y is positive.
Solution
By expanding the equation 7−px−x2=16−(q+x)2, we arrive at the equation 7−px−x2=16−q2−2qx−x2. Since this is true for all x, we can equate coefficients, giving 7=16−q2 and −p=−2q.
Thus q2=9. Since the question tells us that p and q are both positive, we must have q=3, so p=6.
Thus y=7−6x−x2=16−(3+x)2.
As y=16−(3+x)2, we see that the maximum value of y is 16, and this occurs when x=−3.
(This is because (3+x)2≥0 no matter what the value of x, so y is maximised when (3+x)2=0.)
Again, the second representation is useful here: y is positive when 16−(3+x)2>0, that is, when (3+x)2<16. This gives −4<3+x<4, so −7<x<1.
Alternatively, we can use the first representation, y=7−6x−x2=−(x2+6x−7)=−(x−1)(x+7). This is zero when x=1 or x=−7. As this is a “vertex-up” parabola (since the coefficient of x2 is negative), y>0 when −7<x<1.