Complete the following:
(a) Use the Leading Coefficient Test to determine the graph's end behavior.
(b) Find the x-intercepts. State whether the graph crosses the x-axis or touches the x-axis
and turns around at each intercept. Show your work. (c) Find the y-intercept. Show your work.
f(x) = x2(x + 2)
(a).
(b).
(c).
(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.
(a) To use the Leading Coefficient Test to determine the graph's end behavior, we need to look at the leading term of the function.
In this case, the leading term is x^2. Since the degree of the leading term is even, we know that as x approaches positive infinity and negative infinity, the function will also approach positive infinity.
Therefore, the end behavior of the graph is that it will rise towards positive infinity as x approaches both positive and negative infinity.
(b) To find the x-intercepts, we set f(x) equal to zero and solve for x.
f(x) = x^2(x + 2) = 0
Since the product of any factors will be zero if and only if one or more of the factors is zero, we set each factor equal to zero and solve for x.
Setting x^2 = 0 gives us x = 0 (Double root at x = 0).
Setting (x + 2) = 0 gives us x = -2.
Therefore, the x-intercepts are at x = 0 and x = -2.
Now, to determine if the graph crosses the x-axis or touches it and turns around at each intercept, we can look at the multiplicity of each root.
For x = 0, the multiplicity is 2 because x^2 is squared. This means that the graph touches the x-axis and turns around at x = 0.
For x = -2, the multiplicity is 1 because (x + 2) has a power of 1. This means that the graph crosses the x-axis at x = -2.
(c) To find the y-intercept, we set x equal to zero and evaluate f(x).
f(x) = x^2(x + 2)
f(0) = 0^2(0 + 2)
f(0) = 0
Therefore, the y-intercept is at y = 0.
so, what is the leading coefficient test?
f(x) = x^2 (x+2)
if f(x)=0, then you must have either
x^2 = 0
or (x+2) = 0
Note that x^2 is always positive. SO, at x=0, the graph touches the x-axis from above, then goes up again
at x = -2, f(x) changes sign from - to +, so the graph crosses the x-axis.
In general,
f touches at even powers of roots
f crosses at odd powers
I figure you can find y when x=0 for the y-intercept.
This is pretty basic stuff, mostly Algebra I. Better review the pertinent sections in your text.