For the first one,
first thing u need to do is find the values of p and q, although p is not really necessary, but we shall find them anyways
substitute in the values for x and y into
y = p - (x - q)²
At (-1,0)
0 = p - ( -1 - q )²
0 = p - (q² + 2q + 1)
p = q² + 2q + 1
At (5,0)
0 = p - ( 5 - q )²
0 = p - (25 - 10q +q²)
p = q² - 10q + 25
Equate both the p's
q² - 10q + 25 = q² + 2q + 1
Giving,
- 12q = -24
q = 2
q² + 2q + 1 = p
(2)²+2(2) + 1 = p
4 + 4 + 1 = p
p = 9
now we know that
y = 9 - (x - 2)²
to find the maximum values, you need to differentiate
dy/dx = [USING CHAIN RULE]
======> -2(x-2)(1)
=> -2(x-2)
at maximum values, dy/dx = 0
therefore
-2(x-2) = 0
x- 2 = 0
x = 2
y = 9 - (2 - 2)²
y = 9
therefore the coordinates of maximum are (2,9)