## Calculus: Early Transcendentals 8th Edition

We wish to find the point on the line y=2x+3 closest to (0,0). A point on the line is (x,y) or (x, 2x + 3). Using the Distance Formula, we get: D = $\sqrt (2x + 3-0)^2 + (x - 0)^2$ = $\sqrt (2x + 3)^2 + x^2$ = $\sqrt 5x^2 + 12x + 9$ We can simply look at the expression inside the radical. The minimum value of x that gives us the minimum of $5x^2 + 12x + 9$ will also give the minimum for $\sqrt 5x^2 + 12x + 9$ y = $5x^2 + 12x + 9$ Now we find where $y' = 0$ $y' = 10x + 12$ $0 = 10x + 12$ $10x = -12$ $x = -1.2$ = $-6/5$ Since $y = 5x^2 + 12x + 9$ is an upward opening parabola, this is the minimum. Now plug it into the equation to find the y-coordinate: $y = 2(-1.2) + 3$ = 0.6 = 3/5 (-1.2, 0.6)