Explanation:
the distance between 2 points (x1, y1) and (x2, y2) is calculated via Pythagoras over the coordinate differences (as the coordinate differences are the legs and the direct distance is the Hypotenuse of a right-angled triangle) - which is called the distance formula (but it is actually Pythagoras) :
d = sqrt((x1 - x2)² + (y1 - y2)²)
one given point is (-5, 3).
for the second point remember,
f(x) = y
and so,
y = (x - 1)²
which makes the second point (x, (x - 1)²).
and we put this into the distance formula :
d = sqrt((-5 - x)² + (3 - (x - 1)²)²) =
= sqrt(25 + 10x + x² + (3 - (x² - 2x + 1))²) =
= sqrt(25 + 10x + x² + (3 - x² + 2x - 1)²) =
= sqrt(25 + 10x + x² + (-x² + 2x + 2)²) =
= sqrt(25 + 10x + x² + x⁴ - 2x³ - 2x² - 2x³ + 4x² + 4x - 2x² + 4x + 4) =
= sqrt(29 + x⁴ - 4x³ + x² + 18x)
to find the minimum we create the first derivative d' and find its zeros.
remember the chain rule :
the derivative of f(g(x)) is f'(g(x))⋅g'(x).
and remember that the square root of something means the exponent of 1/2 for that something.
so,
d' = 1/2 × (29 + x⁴ - 4x³ + x² + 18x)^-1/2 × (4x³ - 12x² + 2x + 18)
we remember also that a negative exponent means 1/...
d' = (1/(2(29 + x⁴ - 4x³ + x² + 18x)^1/2) × (4x³ - 12x² + 2x + 18)
d' = (4x³ - 12x² + 2x + 18)/(2×sqrt(29 + x⁴ - 4x³ + x² + 18x))
now, d' = 0 for any extreme value (maximum and minimum).
we want to find the minimum as the shortest distance.
a fraction is 0, when the numerator is 0 (and the denominator is not) :
4x³ - 12x² + 2x + 18 = 0
2x³ - 6x² + x + 9 = 0
with a little bit of trying we find x = -1 is a solution.
the corresponding y is then
y = (-1 - 1)² = (-2)² = 4
giving us the point (-1, 4).
since the maximum distance from a point to a quadratic function can only be infinity (twice), the found solution can only be the minimum.
remember, a polynomial of third degree (highest exponent of x is 3) has exactly 3 zeros.
therefore, the point on the curve that is closest to the given point (-5, 3) is
(-1, 4)