The problem asks for the equation of a parabola in vertex form that has the same shape as either $f(x) = 5x^2$ or $g(x) = -5x^2$, but with a given minimum of 0 at $x = 3$.

Key Points:

1. Shape of the parabola: The phrase "same shape" implies that the coefficient of $x^2$ should be the same. Since the graph must have a minimum, it will take the form of $f(x) = 5x^2$ (as the positive sign ensures a parabola opening upwards with a minimum).

2. Vertex form of a parabola: The vertex form of a parabola is $h(x) = a(x - h)^2 + k$ where $(h, k)$ is the vertex of the parabola.

3. Vertex information: The problem gives that the minimum is 0 at $x = 3$, meaning the vertex is at $(3, 0)$. Therefore, $h = 3$ and $k = 0$.

Equation Derivation:

• Using the vertex form $h(x) = a(x - h)^2 + k$,
• The vertex is at $(3, 0)$, so $h(x) = a(x - 3)^2$.
• The parabola has the same shape as $f(x) = 5x^2$, so $a = 5$.

Thus, the equation is: $h(x) = 5(x - 3)^2$

Let me know if you'd like further clarification, or if you have any questions.

Here are 5 related questions:

1. How would the equation change if the minimum was at a different point, such as $x = -2$?
2. What would the equation look like if the parabola had a maximum instead of a minimum?
3. How does the value of $a$ affect the steepness of the parabola?
4. What happens if the coefficient $a$ becomes negative?
5. How do you convert a standard form quadratic equation into vertex form?

Tip: In a quadratic equation, the sign of the leading coefficient $a$ determines whether the parabola opens upwards or downwards.