To find the quadratic function $f(x)$ given its vertex and y-intercept, we'll use the vertex form of a quadratic function and some additional information.

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

Given:

• Vertex: $(-2, 5)$
• Y-intercept: $(0, -3)$

First, plug in the vertex into the vertex form: $f(x) = a(x + 2)^2 + 5$

Next, use the y-intercept to find the value of $a$. The y-intercept means when $x = 0$, $f(x) = -3$. Plug these values into the function: $-3 = a(0 + 2)^2 + 5$ $-3 = a(4) + 5$ $-3 = 4a + 5$ $-3 - 5 = 4a$ $-8 = 4a$ $a = -2$

So the quadratic function is: $f(x) = -2(x + 2)^2 + 5$

Let's expand this to confirm it fits the given conditions: $f(x) = -2(x^2 + 4x + 4) + 5$ $f(x) = -2x^2 - 8x - 8 + 5$ $f(x) = -2x^2 - 8x - 3$

Thus, the quadratic function is: $f(x) = -2x^2 - 8x - 3$

Would you like more details or have any questions? Here are some additional questions you might find interesting:

1. How do you find the vertex of a quadratic function given its standard form?
2. What are the steps to convert a quadratic function from standard form to vertex form?
3. How do you determine the axis of symmetry of a quadratic function?
4. What is the significance of the discriminant in a quadratic equation?
5. Can you derive the quadratic formula from the general form of a quadratic equation?

Tip: To quickly determine the vertex form from the standard form $ax^2 + bx + c$, use the formula for the vertex: $h = -\frac{b}{2a}$ and $k = f(h)$.