The given equation is:

$y = -2x^2 + 160x$

This is a quadratic equation in the form $y = ax^2 + bx + c$, where $a = -2$, $b = 160$, and $c = 0$. Since $a < 0$, the parabola opens downwards, and the vertex represents the maximum value of $y$.

Step 1: Find the $x$-coordinate of the vertex

The formula for the $x$-coordinate of the vertex for a quadratic equation is:

$x = -\frac{b}{2a}$

Substitute $a = -2$ and $b = 160$:

$x = -\frac{160}{2(-2)} = \frac{160}{4} = 40$

So, the $x$-coordinate of the vertex is $x = 40$.

Step 2: Find the maximum value of $y$

Substitute $x = 40$ into the equation $y = -2x^2 + 160x$:

$y = -2(40)^2 + 160(40)$

Calculate step by step:

1. $40^2 = 1600$,
2. $-2(1600) = -3200$,
3. $160(40) = 6400$,
4. $-3200 + 6400 = 3200$.

Thus, the maximum value of $y$ is:

$\boxed{3200}$

Correct answer:

d. 3200

Would you like further clarification or more examples?

Here are 5 related questions to explore further:

1. What is the general formula for the vertex of a quadratic function?
2. How can we determine if a parabola opens upwards or downwards?
3. Can you calculate the minimum value of a quadratic function when $a > 0$?
4. How do you rewrite a quadratic equation in vertex form?
5. How would the solution change if the coefficient of $x^2$ were positive?

Tip: Always double-check the sign of $a$ in a quadratic equation to determine whether the vertex gives a maximum or minimum.