Find a cubic function
y = ax3 + bx2 + cx + d
whose graph has horizontal tangents at the points
well, horizontal tangent,y'=0
y'=0=3ax^2+bx+c
so you solve that quadratic equation
x=(-b+-sqrt(b^2-4ac))/2a
To find a cubic function with horizontal tangents at specific points, we need to find the values of a, b, c, and d that satisfy the given condition.
Let's say we want the graph to have horizontal tangents at the points (p, q) and (r, s).
When a function has a horizontal tangent at a particular point, the derivative of the function at that point is equal to zero.
So, let's start by finding the derivative of the cubic function y = ax^3 + bx^2 + cx + d.
y' = 3ax^2 + 2bx + c
Now, we can set up two equations using the given points (p, q) and (r, s) where the derivative is equal to zero.
For the point (p, q):
0 = 3ap^2 + 2bp + c -- Equation (1)
For the point (r, s):
0 = 3ar^2 + 2br + c -- Equation (2)
We now have a system of two equations with three unknowns (a, b, and c). We need one more equation to solve the system.
To find the third equation, we can use the fact that the cubic function goes through the points (p, q) and (r, s). Let's substitute these values into the cubic function equation.
For the point (p, q):
q = ap^3 + bp^2 + cp + d -- Equation (3)
For the point (r, s):
s = ar^3 + br^2 + cr + d -- Equation (4)
Now, we have a system of three equations with three unknowns (a, b, c).
By solving these three equations simultaneously, we can find the values of a, b, c, and d that satisfy the given condition of horizontal tangents at the points (p, q) and (r, s).
To find a cubic function with horizontal tangents, we need to find the values of a, b, c, and d.
Given that the graph has horizontal tangents at certain points, it means that the slope of the function at those points is zero. In other words, the derivative of the function at those points is zero.
So, let's find the derivative of the function y = ax^3 + bx^2 + cx + d.
Take the derivative of each term with respect to x:
y' = 3ax^2 + 2bx + c
To find the points where the graph has horizontal tangents, we set the derivative equal to zero:
3ax^2 + 2bx + c = 0
Now, we have an equation with three variables: a, b, and c. It is not possible to solve this equation with only one equation, so we need additional information or conditions to determine the values of a, b, and c.
For example, let's say we want the cubic function to have horizontal tangents at the points (1, 2) and (-1, -2). We can substitute these points into the equation:
When x = 1:
3a(1)^2 + 2b(1) + c = 0
3a + 2b + c = 0
When x = -1:
3a(-1)^2 + 2b(-1) + c = 0
3a - 2b + c = 0
Now we have a system of two equations with three variables (a, b, c). We can solve this system of equations to find the values of a, b, and c.
Once we have the values of a, b, and c, we can substitute them back into the original equation y = ax^3 + bx^2 + cx + d to find the value of d.
Keep in mind that the values of a, b, c, and d may not be unique, as there are infinite possible cubic functions with horizontal tangents at certain points.