The area of the triangle formed by x- and y-intercepts of the parabola y=5(x-3)(x+k) is equal to 1.5 square units. Find all possible values of k.
x =3 is one of the zeroes....this would be one of the legs of the triangle....the other would occur on the y axis and needs to be = |1| for the area (1/2 b*h) = 1.5
the y axis crossing occurs when x = 0
-1 = 5 (x-3)(x+k) sub in x = 0
-1/5 = (-3)(k)
k = 1/15
when y = +1 1 = 5 (0-3((0+k )
1/5 = - 3k
k = - 1/15
Here is a desmos graph: