Question

The point on the curve $ y = x^2 + 4x + 3 $ that is closest to the line $ y = 3x + 2 $ is:

• A. \( \left( \frac{1}{2}, \frac{5}{4} \right) \)
• B. \( \left( -\frac{1}{2}, \frac{5}{4} \right) \)
• C. \( \left( 2, -\frac{5}{3} \right) \)
• D. \( \left( 2, \frac{5}{3} \right) \)

Hint:

To minimize distance from a curve to a line, substitute into distance formula and minimize the square.

Answer 1

The Correct Option is B

We are given the curve: \[ y = x^2 + 4x + 3 \] And the line: \[ y = 3x + 2 \] To find the point on the curve closest to the line, we minimize the square of the perpendicular distance from a point \( (x, y = x^2 + 4x + 3) \) to the line: \[ \text{Distance} = \frac{|3x - y + 2|}{\sqrt{10}} \Rightarrow \text{Minimize } D^2 = \left( \frac{3x - y + 2}{\sqrt{10}} \right)^2 \] Substitute \( y = x^2 + 4x + 3 \) into the expression: \[ 3x - (x^2 + 4x + 3) + 2 = 3x - x^2 - 4x - 3 + 2 = -x^2 - x - 1 \Rightarrow D^2 \propto (-x^2 - x - 1)^2 \] Now minimize: \[ f(x) = (-x^2 - x - 1)^2 \Rightarrow f(x) = (x^2 + x + 1)^2 \] Differentiate: \[ f'(x) = 2(x^2 + x + 1)(2x + 1) \Rightarrow f'(x) = 0 \Rightarrow x = -\frac{1}{2} \] Find \( y = (-\frac{1}{2})^2 + 4(-\frac{1}{2}) + 3 = \frac{1}{4} - 2 + 3 = \frac{5}{4} \) Thus, the point is \( \boxed{\left( -\frac{1}{2}, \frac{5}{4} \right)} \)