Find the equation of a quartic polynomial whose graph is symmetric about the y-axis and has local maxima at (−3,3) and (3,3) and a y-intercept of 2.
you know that y' = ax(x-3)(x+3) = a(x^3-9x)
That makes
y = a(1/4 x^4 - 9/2 x^2) + C
Since y(0) = 2, C=2 and you have
y = ax^2(1/4 x^2 - 9/2) + 2
Since y(3)=3, you have
9a(9/4 * 9 - 9/2) + 2 = 3
a = -4/81
and so you have
y = -4/81 x^2 (1/4 x^2 - 9/2) + 2
see the graph and its properties at
https://www.wolframalpha.com/input/?i=-4%2F81+x%5E2+%281%2F4+x%5E2+-+9%2F2%29+%2B+2
thank you so much! :)
To find the equation of a quartic polynomial satisfying the given conditions, we'll start by considering the symmetry of the graph. If the graph is symmetric about the y-axis, it means the polynomial has only even powers.
Let's assume the general form of a quartic polynomial:
f(x) = ax^4 + bx^3 + cx^2 + dx + e
Since the graph is symmetric about the y-axis, the polynomial satisfies:
f(x) = f(-x)
Substituting x = -x in the general form, we have:
ax^4 - bx^3 + cx^2 - dx + e = ax^4 + bx^3 + cx^2 + dx + e
This reduces to:
-bx^3 - dx = bx^3 + dx
-2bx^3 - 2dx = 0
2bx^3 + 2dx = 0
Since this equation must hold for all x, the coefficients of x^3 and x must both be zero:
2b = 0 and 2d = 0
b = 0 and d = 0
Now, the equation becomes:
f(x) = ax^4 + cx^2 + e
Next, we'll use the given information about the local maxima at (−3,3) and (3,3) along with the y-intercept at (0,2) to determine the values of a, c, and e.
At x = -3, f(-3) = 3:
3 = a(-3)^4 + c(-3)^2 + e
3 = 81a + 9c + e
At x = 3, f(3) = 3:
3 = a(3)^4 + c(3)^2 + e
3 = 81a + 9c + e
At x = 0, f(0) = 2:
2 = a(0)^4 + c(0)^2 + e
2 = e
We now have a system of three equations:
3 = 81a + 9c + e
3 = 81a + 9c + e
2 = e
Simplifying this system, we get:
81a + 9c + 2 = 3
81a + 9c + 2 = 3
2 = e
From the last equation, e = 2. Substituting this back into the first two equations, we find that the variables are eliminated:
81a + 9c = 1
81a + 9c = 1
These equations are identical, indicating that there are infinitely many solutions. We can choose any values for a and c as long as they satisfy the equation 81a + 9c = 1. For example, let's choose a = 1 and c = 0, then the equation becomes:
81(1) + 9(0) = 1
81 = 1
Therefore, the equation of a quartic polynomial satisfying the given conditions is:
f(x) = x^4 + 2