what point on the line y = 3x + 6 is closest to the origin? let d be the distance between the two points…

what point on the line y = 3x + 6 is closest to the origin? let d be the distance between the two points. what is the objective function in terms of the d = √(10x² + 36x + 36) (type an expression.) the interval of interest of the function is (-∞,∞). (simplify your answer. type your answer in interval notation.) the point on the line y = 3x + 6 closest to the origin is . (type an ordered pair. simplify your answer. type integers or simplified fractions.)

what point on the line y = 3x + 6 is closest to the origin? let d be the distance between the two points. what is the objective function in terms of the d = √(10x² + 36x + 36) (type an expression.) the interval of interest of the function is (-∞,∞). (simplify your answer. type your answer in interval notation.) the point on the line y = 3x + 6 closest to the origin is . (type an ordered pair. simplify your answer. type integers or simplified fractions.)

Answer

Explanation:

Step1: Minimize the distance function

To find the minimum of $D = \sqrt{10x^{2}+36x + 36}$, we can minimize the function $f(x)=10x^{2}+36x + 36$ (since minimizing $f(x)$ is equivalent to minimizing $D$ as the square - root function is monotonic). We take the derivative of $f(x)$ using the power rule. The derivative $f^\prime(x)=\frac{d}{dx}(10x^{2}+36x + 36)=20x+36$.

Step2: Find the critical points

Set $f^\prime(x) = 0$. So, $20x+36 = 0$. Solving for $x$ gives $20x=-36$, then $x=-\frac{36}{20}=-\frac{9}{5}$.

Step3: Find the y - coordinate

Substitute $x = -\frac{9}{5}$ into the line equation $y = 3x+6$. Then $y=3\times(-\frac{9}{5})+6=\frac{-27 + 30}{5}=\frac{3}{5}$.

Answer:

$(-\frac{9}{5},\frac{3}{5})$