(a) The Leading Coefficient Test states that for a polynomial function of the form f(x) = ax^n, where a is the leading coefficient and n is the degree of the polynomial, the end behavior of the graph is determined by the sign of the leading coefficient and the parity of the degree.
In this case, the function f(x) = x^2(x + 2) has a leading coefficient of 1 (since it's not explicitly stated). The degree of the polynomial is 3 (by adding the exponents). Since the degree is odd and the leading coefficient is positive, the end behavior of the graph is as follows:
- As x approaches negative infinity, f(x) approaches negative infinity.
- As x approaches positive infinity, f(x) approaches positive infinity.
(b) To find the x-intercepts, we set f(x) = 0 and solve for x.
Setting f(x) = x^2(x + 2) = 0:
x^2 = 0 or x + 2 = 0
For x^2 = 0, taking the square root of both sides gives us x = 0.
For x + 2 = 0, subtracting 2 from both sides gives us x = -2.
So, the x-intercepts are x = 0 and x = -2. To determine whether the graph crosses or touches the x-axis and turns around at each intercept, we can examine the multiplicity of each intercept.
For x = 0, the multiplicity is 2 since there is a double root (x^2). This means that the graph touches the x-axis but does not cross it at x = 0.
For x = -2, the multiplicity is 1 (x + 2). This means that the graph crosses the x-axis and turns around at x = -2.
(c) To find the y-intercept, we set x = 0 in the function f(x) = x^2(x + 2).
Setting x = 0, we get f(0) = 0^2(0 + 2) = 0.
So, the y-intercept is 0.