The given function is:

$f(x) = -2x^2 + 8x$

This is a quadratic function. Let’s break down some key characteristics of this function:

1. General form: This is in the standard quadratic form $f(x) = ax^2 + bx + c$, where $a = -2$, $b = 8$, and $c = 0$.

2. Vertex: The vertex of a quadratic function in standard form can be found using the formula for the x-coordinate of the vertex:

$x_{\text{vertex}} = -\frac{b}{2a}$

Substituting the values of $a = -2$ and $b = 8$:

$x_{\text{vertex}} = -\frac{8}{2(-2)} = 2$

Now, plug $x = 2$ back into the function to find the y-coordinate of the vertex:

$f(2) = -2(2)^2 + 8(2) = -8 + 16 = 8$

So, the vertex is at $(2, 8)$.

3. Axis of symmetry: The axis of symmetry is the vertical line that passes through the vertex, given by $x = 2$.

4. Direction of opening: Since the coefficient of $x^2$ (which is $-2$) is negative, the parabola opens downward.

5. Y-intercept: The y-intercept occurs when $x = 0$. Substituting $x = 0$ into the function:

$f(0) = -2(0)^2 + 8(0) = 0$

So, the y-intercept is at $(0, 0)$.

Summary:

• Vertex: $(2, 8)$
• Axis of symmetry: $x = 2$
• Direction: Opens downward
• Y-intercept: $(0, 0)$

Would you like more details, or do you have any specific questions?

Here are five related questions:

1. What are the x-intercepts (roots) of the function $f(x) = -2x^2 + 8x$?
2. How do you find the domain and range of a quadratic function?
3. How do transformations affect the graph of a quadratic function like this one?
4. What is the effect of changing the coefficient of $x^2$ on the shape of the parabola?
5. How do you complete the square for $f(x) = -2x^2 + 8x$?

Tip: When solving quadratic equations, always check if you can factorize the expression before using more complex methods like the quadratic formula.