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 x - coordinate? d = √(10x² + 36x + 36) (type an expression.) the interval of interest of the function is . (simplify your answer. type your answer in interval notation.)

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 x - coordinate? d = √(10x² + 36x + 36) (type an expression.) the interval of interest of the function is . (simplify your answer. type your answer in interval notation.)

Answer

Explanation:

Step1: Recall distance formula

The distance $D$ between a point $(x,y)$ and the origin $(0,0)$ is $D=\sqrt{(x - 0)^2+(y - 0)^2}=\sqrt{x^{2}+y^{2}}$. Since $y = 3x+6$, we substitute $y$ into the distance - formula: $D=\sqrt{x^{2}+(3x + 6)^{2}}=\sqrt{x^{2}+9x^{2}+36x + 36}=\sqrt{10x^{2}+36x + 36}$.

Step2: Find the interval of interest

The function $y = 3x+6$ is a linear function, and $x$ can take any real - valued number. So the interval of interest for the function $D(x)=\sqrt{10x^{2}+36x + 36}$ is $(-\infty,\infty)$.

Answer:

$(-\infty,\infty)$