Find the equation of a quartic polynomial whose graph is symmetric about the y-axis and has local maxima at (−1,1) and (1,1) and a y-intercept of -1.

To find the equation of a quartic polynomial with the given properties, we need to consider the symmetry about the y-axis as well as the local maxima and y-intercept.

Since the graph is symmetric about the y-axis, the equation of the quartic polynomial can be written in the form:

f(x) = ax⁴ + bx³ + cx² + dx + e

Since the local maxima occur at (-1,1) and (1,1), these points must be critical points of the quartic polynomial. The derivative of f(x) with respect to x will give us the necessary information to proceed.

Differentiating f(x) with respect to x:
f'(x) = 4ax³ + 3bx² + 2cx + d

At x = -1, f'(-1) = 0
So, substituting x = -1 into f'(x) gives us:
0 = 4a(-1)³ + 3b(-1)² + 2c(-1) + d
0 = -4a + 3b - 2c + d --(1)

Similarly, at x = 1, f'(1) = 0
So, substituting x = 1 into f'(x) gives us:
0 = 4a(1)³ + 3b(1)² + 2c(1) + d
0 = 4a + 3b + 2c + d --(2)

Given that the y-intercept is -1, we can also substitute x = 0 into f(x):
f(0) = a(0)⁴ + b(0)³ + c(0)² + d(0) + e
-1 = e --(3)

Now we have a system of three equations (1), (2), and (3) in four unknowns a, b, c, and d. To solve this system, we need one more equation. Since the graph is symmetric about the y-axis, the quartic polynomial will have an even degree, and therefore the coefficient of the x-term (d) will be zero. This gives us the fourth equation:

d = 0

Now we can substitute d = 0 into equations (1) and (2):

0 = -4a + 3b - 2c --(4)
0 = 4a + 3b + 2c --(5)

By adding equations (4) and (5), we can eliminate a:

0 = 3b -> b = 0

Substituting b = 0 into equations (4) and (5) gives us:

0 = -4a - 2c --(6)
0 = 4a + 2c --(7)

Adding equations (6) and (7), we can eliminate c:

0 = 0 -> True

This means that a and c can have arbitrary values. Let's assume a = 1 and c = 1 for simplicity.

Substituting a = 1, b = 0, c = 1, and d = 0 into equation (3), we find e = -1.

Therefore, the equation of the quartic polynomial is:

f(x) = x⁴ + x² - 1.

To find the equation of a quartic polynomial with these characteristics, we can start by considering the general form of a quartic polynomial:

f(x) = ax^4 + bx^3 + cx^2 + dx + e

Considering that the graph is symmetric about the y-axis, we know that the polynomial only contains even-degree terms. So, we can simplify it to:

f(x) = ax^4 + cx^2 + e

Now, let's apply the given information to determine the values of a, c, and e.

1. Local maxima at (-1,1) and (1,1):
For a polynomial, the coordinates of the local maxima occur at critical points where the derivative equals zero. Since the local maxima have the same y-coordinate (1), we can conclude that the derivative equals zero at x = -1 and x = 1.

Taking the derivative of f(x), we have:

f'(x) = 4ax^3 + 2cx

Setting this derivative equal to zero, we get two equations:

4a(-1)^3 + 2c(-1) = 0 ---> -4a - 2c = 0 (equation 1)
4a(1)^3 + 2c(1) = 0 ---> 4a + 2c = 0 (equation 2)

Simplifying equation 1 and equation 2, we find:

-4a - 2c = 0 ---> -2a - c = 0 (equation 3)
4a + 2c = 0 ---> 2a + c = 0 (equation 4)

Adding equation 3 and equation 4 together, we obtain:

0 = 0

This indicates that equations 3 and 4 are equivalent, meaning they represent the same line. Therefore, we conclude that there is no restriction on the values of a and c. Let's choose arbitrary values to proceed.

Let a = 1 and c = 1, which satisfies equation 3 and equation 4.

2. Y-intercept at (-1,1):
To find the value of e, we can substitute x = 0 into the polynomial equation f(x):

f(0) = a(0)^4 + c(0)^2 + e
f(0) = e

Given that the y-intercept is -1, we have:

f(0) = -1
e = -1

Substituting a = 1, c = 1, and e = -1 into the general quartic polynomial equation, we get:

f(x) = x^4 + x^2 - 1

Therefore, the equation of the quartic polynomial that satisfies the given conditions is f(x) = x^4 + x^2 - 1.

y = a x^4 + b x^3 + c x^2 + d x + e

when x = 0, y = -1 so e = -1
so
y = a x^4 + b x^3 + c x^2 + d x - 1

for max or min
dy/dx = 0 = 4 a x^3 + 3 b x^2 + +2 c x + d

when x = -1
0 = 4 a (-1) +3b(1)+2c(-1)+d
1 = a (1) + b(-1) + c(1) +d(-1) - 1

when x = +1
0 = 4a + 3 b + 2c + d
1 = a+b+c+d-1
--------------------------------
start solving
0 = -4a+3b-2c+d
0 = +4a+3b+2c+d
-----------------
0 = 6b +2d

2 = a-b+c-d
2 = a+b+c+d
-------------
0 = -2b-2d

0 = 6b +2d
0 = -2b-2d
-----------
8 b = 0
b=0
then d = 0
(in fact from symmetry I bet all odd terms are 0)
now go back and get a and c