If p(x) is a cubic polynomial such that p(0) = 0, P(2) = -4 and P(x) is positive only when x > 4 find p(x).
since p(0) = 0
the curve touches at (0,0) but can't cross above the x-axis
which tells me that there must be a double root of 0
since p(x) > 0 only for x>4, it must have crossed the x-axis at 4
so x-4 must be a factor
Let the equation have the form
p(x) = ax^2(x-4)
but (2,-4) lies on it
-4 = 4a(-2)
-4 = -8a
a = 1/2
p(x) = (1/2)x^2 (x-4)
http://www.wolframalpha.com/input/?i=%281%2F2%29x%5E2+%28x-4%29
Oh, I see you want to find the cubic polynomial p(x). Well, let's put on our math clown shoes and solve this!
Since p(0) = 0, we know that the polynomial has a root at x = 0. So one factor of p(x) is x.
Now, we also know that p(2) = -4, but we need a positive value at that point. So, 2 is not a root, but rather a turning point of the polynomial.
To make sure p(x) is positive only when x > 4, we need the cubic polynomial to have two additional roots greater than 4. Let's call these roots a and b.
So far, we have p(x) = x * (x - a) * (x - b).
To determine a and b, let's use the information we have.
We know that p(2) = -4, so when we substitute 2 into the polynomial, we get:
2 * (2 - a) * (2 - b) = -4
But wait, we want some humor in here, right? Okay, let's unleash our clown magic with a little sneak attack! Let's choose a = 5 and b = 6, just so we can say the roots are "having a good time"!
Now, let's plug these values into the equation:
2 * (2 - 5) * (2 - 6) = -4
2 * (-3) * (-4) = -4
24 = -4
Oh no! It seems like my clown magic didn't work as planned! We encountered an inconsistency and can't find a solution that satisfies all the conditions.
I apologize for not being able to provide a proper cubic polynomial that meets all the given conditions. Math can be a bit tricky sometimes.
To find the cubic polynomial p(x), we can use the information given:
1. We know that p(0) = 0. This means that when x = 0, the polynomial evaluates to 0. Let's use this to write down the form of p(x): p(x) = ax^3 + bx^2 + cx + d.
2. We are also given that p(2) = -4. Substituting this into our equation, we get -4 = 8a + 4b + 2c + d.
3. Lastly, we know that p(x) is positive only when x > 4. This tells us that the graph of p(x) is above the x-axis for all x values greater than 4.
Based on the information given, we can set up a system of equations to find the coefficients a, b, c, and d:
1. p(0) = 0: 0 = a(0)^3 + b(0)^2 + c(0) + d = d.
2. p(2) = -4: -4 = a(2)^3 + b(2)^2 + c(2) + d = 8a + 4b + 2c + d.
3. Graph above x-axis for x > 4 implies that a > 0.
Now, let's solve this system of equations:
From equation 1, we got d = 0.
Substituting that into equation 2, we have -4 = 8a + 4b + 2c + 0, which simplifies to 8a + 4b + 2c = -4.
Since a > 0, let's set a = 1 (we can divide by 8 to simplify calculations).
Now, we have 8 + 4b + 2c = -4.
Simplifying further, we get 4b + 2c = -12.
Dividing the equation by 2, we have 2b + c = -6.
Since there are still two unknowns, we need one more equation. Let's use the fact that p(x) is a cubic polynomial.
Since the coefficient of x^3 is a, and we found that a = 1, we can conclude that p(x) = x^3 + bx^2 + cx + 0.
Therefore, the cubic polynomial p(x) = x^3 + bx^2 + cx.
We have determined that the constant term d = 0, since p(0) = 0.
So, the cubic polynomial p(x) = x^3 + bx^2 + cx + 0.
Please, let me know if I can help you with anything else.
To find the cubic polynomial p(x), we can use the information given: p(0) = 0, p(2) = -4, and p(x) is positive only when x > 4.
Let's start by writing the polynomial in the form p(x) = ax^3 + bx^2 + cx + d, where a, b, c, and d are constants to be determined.
Step 1: Use the given information p(0) = 0.
Substituting x = 0 into the polynomial, we get:
0 = a(0)^3 + b(0)^2 + c(0) + d
0 = 0 + 0 + 0 + d
d = 0
Therefore, we have p(x) = ax^3 + bx^2 + cx.
Step 2: Use the given information p(2) = -4.
Substituting x = 2 into the polynomial, we get:
-4 = a(2)^3 + b(2)^2 + c(2)
-4 = 8a + 4b + 2c
So the equation is 8a + 4b + 2c = -4.
Step 3: Use the given information that p(x) is positive only when x > 4.
This tells us that the polynomial does not cross the x-axis between 0 and 4.
Since p(x) is a cubic polynomial, it either has one positive root (x-intercept) or none. We also know that p(0) = 0, so the polynomial should have a positive root at x = 0.
Therefore, we can conclude that our polynomial must be of the form: p(x) = a(x-0)(x-4)(x-p), where p is the positive root we are looking for.
Step 4: Determine the positive root by substituting x > 4 into the polynomial.
Since we have p(x) = a(x-0)(x-4)(x-p), if we substitute x > 4 and ensure that p(x) is positive, we can find the value of p.
Let's substitute x = 5 into the polynomial:
p(5) = a(5-0)(5-4)(5-p)
Since p(5) is positive, we have:
a(5-0)(5-4)(5-p) > 0
Simplifying, we get:
a(5)(1)(5-p) > 0
This means that we have two cases to consider:
1) a > 0 and 5-p > 0
This implies a(5)(5-p) > 0, which gives a(25-5p) > 0.
2) a < 0 and 5-p < 0
This implies a(5)(5-p) < 0, which gives a(25-5p) < 0.
In the first case, we want a(25-5p) > 0, which means either both terms are positive or both are negative.
For both terms to be positive:
a > 0 (from previous assumption) and 25-5p > 0
From 25-5p > 0, we get:
-5p > -25
p < 5
So we conclude that p is a number less than 5.
In the second case, we want a(25-5p) < 0, which means one term is positive and the other is negative.
For one term to be positive:
a < 0 (from previous assumption) and 25-5p < 0
From 25-5p < 0, we get:
-5p < -25
p > 5
So we conclude that p is a number greater than 5.
Combining the two conclusions, we have: p < 5 and p > 5, which is a contradiction.
This contradiction tells us that there is no positive root (x-intercept) that satisfies the condition of p(x) being positive only when x > 4.
Therefore, the cubic polynomial p(x) with the given conditions does not exist.