Precalculus Question

Question

Which point on the curve y= square root of x-2 and then plus 1 is closest to the point (4,1)? What is the minimum distance?

Answer 1

Which point on the curve y= square root of x-2 and then plus 1 is closest to the point (4,1)? What is the minimum distance?

So we need the distance between the points (4,1) and [x,(x-2)^1/2+1] noting x≥2

D={(x-4)²+[(x-2)^1/2+1-1]²}^1/2

D={(x-4)²+[(x-2)^1/2]²}^1/2

D={(x-4)²+(x-2)}^1/2

D={x²-8x+16+x-2}^1/2

D={x²-7x+14}^1/2

The quadratic x²-7x+14 is concave up because the coefficient of the squared term is positive.

Compare x²-7x+14 to ax²+bx+c and calculate b²-4ac to determine the kind of roots:

Recall

If b²-4ac=0, there is a double real root.

If b²-4ac<0, there are no real roots

If b²-4ac>0, there are two real roots

Here b²-4ac=(-7)2-4(1)(14)=49-56<0 so there are no real roots

We know if y=x²-7x+14≥0

y=x²-7x+14=(x-3.5)²-3.5²+14

To find the minimum (vertex) we have to put the quadratic into the general form of a parabola 4p(y-k)=(x-h)² where (h,k) is the vertex (extreme point).

y=(x-3.5)²+1.75

y-1.75=(x-3.5)²

Compare to the general form of the parabola and

h=3.5

k=1.75

4p=1

p=1/4

This is a concave up parabola with its vertex (minimum) at (h,k)=(3.5,1.75).

Since y=x²-7x+14 has its minimum at (3.5,1.75), then D is minimum at that point (D=0).

Solution: the point closest to (4,1) is (3.5,1.75) and the minimum distance is zero.