To find a quadratic function of the form $y = ax^2$ that passes through the point $(-4, -32)$, we can substitute the point's coordinates into the equation and solve for the coefficient $a$.

Given:

• Point: $(-4, -32)$
• Function: $y = ax^2$

Substitute $x = -4$ and $y = -32$ into the equation:

$-32 = a(-4)^2$

Now, simplify the expression:

$-32 = a \cdot 16$

Solving for $a$, we get:

$a = \frac{-32}{16} = -2$

Therefore, the quadratic function is:

$y = -2x^2$

Final Answer:

The quadratic function that passes through the point $(-4, -32)$ is:

$\boxed{y = -2x^2}$

Would you like to see a graph of this function or further details on quadratic functions?

Related Questions:

1. How do you find a quadratic function that passes through three points?
2. What are the vertex and axis of symmetry for the quadratic function $y = -2x^2$?
3. How does the coefficient $a$ affect the shape of a quadratic function?
4. What is the process for finding the roots of a quadratic equation like $y = -2x^2$?
5. Can you derive the standard form of a quadratic equation from its vertex form?

Tip:

To identify how a quadratic function behaves, always look at the sign and magnitude of the coefficient $a$; a negative $a$ inverts the parabola, while its absolute value determines the "width" or "steepness."