To construct a rational function with a slant asymptote and x-intercepts at -1 and 4, we can start by considering the x-intercepts.
Given that the x-intercepts are -1 and 4, the factors in the denominator will be (x + 1) and (x - 4).
Now, to introduce a slant asymptote, we need the degree of the numerator to be greater than the degree of the denominator by 1. In this case, we will make the numerator have a degree of 2 and the denominator a degree of 1.
Let's assume the rational function is:
f(x) = (ax^2 + bx + c) / (x + 1)(x - 4)
To find the values of a, b, and c, we can use the x-intercepts.
When x = -1, the function should equal zero:
0 = (a(-1)^2 + b(-1) + c) / (-1 + 1)(-1 - 4)
0 = (a - b + c) / (-1)(-5)
0 = (a - b + c) / 5
This gives us the equation: a - b + c = 0
When x = 4, the function should equal zero:
0 = (a(4)^2 + b(4) + c) / (4 + 1)(4 - 4)
0 = (16a + 4b + c) / (5)(0)
0 = 16a + 4b + c
This gives us the equation: 16a + 4b + c = 0
By solving the system of equations formed by the two equations above, we can find the values of a, b, and c.
With this information, we can construct the rational function with a slant asymptote and x-intercepts at -1 and 4.