Find the point on the parabola y=3x^2+4x-8 that is closest to the point (-2,-3)

Answer 1

point on the parabola y=3x^2+4x-8 that is closest to the point (-2,-3):

The distance^2 from the point to the parabola = (x -(-2))^2 + (y-(-3))^2 where y = 3x^2+4x-8 is

(x + 2)^2 + (3x^2+4x-8 + 3)^2

(finding the x value that minimizes the distance squared gives the same result as if we were to minimize distance itself.)

this is minimized when the derivative of d^2 with respect to x, d(d^2)/dx, is 0

2(x+2) + 2(3x^2+4x-5)(6x + 4) = 0 or

x+2 + (3x^2+4x-5)(6x + 4) = 0

18x^3 + 36x^2 − 13x − 18 = 0

Using a calculator, we get three values for x, x₁ = 0.74868, x₂ = -2.11808, x₃ = -0.63061,

and plugging back into the formula for d^2, shows that x₂ = -2.11808 produces the minimum distance.

evaluating the original function at x₂ = -2.11808, gives y = -3.013531. So it looks like the closest point on the parabola to (-2, -3) is (-2.11808, -3.013531). Maybe we could solve the cubic derivative algebraically, but for the sake of speed of generating a free solution, I assumed that you could use a calculator.