which points on the graph of y = 6 - x² are closest to the point (0, 3)?\n\na. (-√5/2, 19/4) and (√5/2…

Explanation:

Step1: Define the distance formula

Let a point on the graph $y = 6 - x^{2}$ be $(x,y)=(x,6 - x^{2})$. The distance $d$ between $(x,6 - x^{2})$ and $(0,3)$ is given by the distance formula $d=\sqrt{(x - 0)^{2}+((6 - x^{2})-3)^{2}}=\sqrt{x^{2}+(3 - x^{2})^{2}}$. To simplify calculations, we can minimize the square of the distance function $D(x)=x^{2}+(3 - x^{2})^{2}=x^{2}+9 - 6x^{2}+x^{4}=x^{4}-5x^{2}+9$.

Step2: Find the derivative of $D(x)$

Using the power - rule, $D^\prime(x)=4x^{3}-10x = 2x(2x^{2}-5)$.

Step3: Set the derivative equal to zero

Set $D^\prime(x)=0$. Then $2x(2x^{2}-5)=0$. This gives us two cases:

• Case 1: $2x = 0$, so $x = 0$.
• Case 2: $2x^{2}-5=0$, then $x^{2}=\frac{5}{2}$, so $x=\pm\frac{\sqrt{10}}{2}$.

Step4: Find the corresponding $y$ - values

When $x=\pm\frac{\sqrt{10}}{2}$, $y = 6 - x^{2}=6-\frac{5}{2}=\frac{7}{2}$.

Answer:

B. $(-\frac{\sqrt{10}}{2},\frac{7}{2})$ and $(\frac{\sqrt{10}}{2},\frac{7}{2})$