If we want to find the points on a graph that are closest to a given point, we need to find the minimum distance between that point and the function. We can use calculus to find this minimum.
Let's start by defining the function:
f(x) = 4-x²
We also define its derivative (which represents the slope of the function at any point x):
f'(x) = -2x
The square of the distance between any point (x, y) on the map and the point (0, 2) is:
D² = x² + (f(x) - 2)².
We are looking for the x-value that minimizes D². To find the minimum of a function, we take the derivative and set it equal to zero.
The derivative of D² is:
D'² = 2 * x + 2 * (f(x) - 2) * f'(x)
Setting this equal to zero gives us the equation:
2 * x + 2 * (f(x) - 2) * f'(x) = 0.
We solve this equation for x to get the x-values for which D² is minimized. We get three solutions:
x = 0,
x = -sqrt(6)/2,
x = sqrt(6)/2.
We substitute these values of x into the function f(x) to get the corresponding y-values:
for x = 0, we get y = f(0) = 4,
for x = -sqrt(6)/2, we get y = f(-sqrt(6)/2) = 5/2,
for x = sqrt(6)/2, we get y = f(sqrt(6)/2) = 5/2.
So, the points on the graph that are closest to the point (0, 2) are (0, 4), (-sqrt(6)/2, 5/2), and (sqrt(6)/2, 5/2).