We are tasked with finding the points on the graph of the function $f(x) = x^2$ that are closest to the point $(0, 9)$. Here's the step-by-step solution to solve this optimization problem.

Step 1: Define the Distance Formula

The distance between a point $(x, y)$ on the curve and the given point $(0, 9)$ is given by the distance formula:

$d = \sqrt{(x - 0)^2 + (y - 9)^2}$

Since $y = x^2$ (from the equation of the curve), we substitute this into the distance formula:

$d = \sqrt{x^2 + (x^2 - 9)^2}$

This simplifies to:

$d = \sqrt{x^2 + (x^4 - 18x^2 + 81)}$

$d = \sqrt{x^4 - 17x^2 + 81}$

Step 2: Minimize the Distance

To minimize the distance $d$, we can minimize $d^2$, as it simplifies the process. Let:

$D(x) = x^4 - 17x^2 + 81$

We will now minimize $D(x)$.

Step 3: Take the Derivative of $D(x)$

Differentiate $D(x)$ with respect to $x$:

$D'(x) = 4x^3 - 34x$

Step 4: Set the Derivative Equal to Zero

Set $D'(x) = 0$ to find the critical points:

$4x^3 - 34x = 0$

Factor the equation:

$2x(2x^2 - 17) = 0$

This gives us two solutions:

1. $x = 0$
2. $2x^2 - 17 = 0$

Solve for $x^2$ in the second equation:

$x^2 = \frac{17}{2}$

Thus, $x = \pm \sqrt{\frac{17}{2}} = \pm \frac{\sqrt{34}}{2}$.

Step 5: Find the Corresponding $y$-values

For $x = 0$:

$y = f(0) = 0^2 = 0$

For $x = \pm \frac{\sqrt{34}}{2}$:

$y = f\left(\pm \frac{\sqrt{34}}{2}\right) = \left(\pm \frac{\sqrt{34}}{2}\right)^2 = \frac{34}{4} = \frac{17}{2}$

Step 6: Points on the Graph

• For the smaller $x$-value $x = -\frac{\sqrt{34}}{2}$, the point is $\left(-\frac{\sqrt{34}}{2}, \frac{17}{2}\right)$.
• For the larger $x$-value $x = \frac{\sqrt{34}}{2}$, the point is $\left(\frac{\sqrt{34}}{2}, \frac{17}{2}\right)$.

Thus, the points on the graph closest to $(0, 9)$ are:

• Smaller $x$-value: $\left(-\frac{\sqrt{34}}{2}, \frac{17}{2}\right)$
• Larger $x$-value: $\left(\frac{\sqrt{34}}{2}, \frac{17}{2}\right)$

Would you like further details or have any questions?

5 Relative Questions:

1. How do we derive the critical points when minimizing a function?
2. Can you explain why minimizing $d^2$ is easier than minimizing $d$?
3. What happens when a point on the graph lies directly above or below the given point?
4. How does the distance formula change if the point moves along the y-axis?
5. Can we always assume symmetry when solving optimization problems on parabolas?

Tip:

When dealing with distance minimization problems, always simplify the square of the distance to avoid working with square roots until the final step.