Let's go step by step to solve the given quadratic function problem. The quadratic function provided is:

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

(a) Express $f(x)$ in vertex form

The vertex form of a quadratic function is:

$f(x) = a(x - h)^2 + k$

Where $(h, k)$ is the vertex of the parabola. To convert the given equation to vertex form, we will complete the square.

1. Factor out the coefficient of $x^2$ from the quadratic and linear terms:

$f(x) = -(x^2 - 16x)$

2. Complete the square:
   • Take the coefficient of $x$, divide it by 2, and square it. In this case: $\left(\frac{-16}{2}\right)^2 = (-8)^2 = 64$
   • Add and subtract 64 inside the parentheses:

$f(x) = -(x^2 - 16x + 64 - 64)$ $f(x) = -( (x - 8)^2 - 64)$

3. Simplify:
   • Distribute the negative sign:

$f(x) = -(x - 8)^2 + 64$

Thus, the vertex form of the function is:

$f(x) = -(x - 8)^2 + 64$

(b) Find the vertex and intercepts

Vertex:

From the vertex form $f(x) = -(x - 8)^2 + 64$, the vertex $(h, k)$ is $(8, 64)$.

Vertex: $(x, y) = (8, 64)$

X-intercepts:

To find the x-intercepts, set $f(x) = 0$ and solve for $x$:

$-x^2 + 16x = 0$ Factor the equation:

$-x(x - 16) = 0$

So the x-intercepts are $x = 0$ and $x = 16$.

X-intercepts: $(0, 0), \quad (16, 0)$

Y-intercept:

The y-intercept occurs when $x = 0$. Substituting $x = 0$ into the original equation:

$f(0) = -(0)^2 + 16(0) = 0$

Y-intercept: $(0, 0)$

(c) Sketch a graph of $f(x)$

• The parabola opens downward because the coefficient of $x^2$ is negative.
• The vertex is $(8, 64)$.
• The x-intercepts are at $(0, 0)$ and $(16, 0)$.
• The y-intercept is at $(0, 0)$.

The parabola will be symmetric about $x = 8$, and the graph should be a downward-facing parabola with the vertex at $(8, 64)$.

(d) Domain and range

• The domain of any quadratic function is all real numbers: $\text{Domain: } (-\infty, \infty)$
• The range is determined by the vertex and the direction the parabola opens. Since the parabola opens downward and the maximum value is 64 (at the vertex), the range is: $\text{Range: } (-\infty, 64]$

Final Answers:

• (a) Vertex form: $f(x) = -(x - 8)^2 + 64$
• (b)
  • Vertex: $(8, 64)$
  • X-intercepts: $(0, 0)$ and $(16, 0)$
  • Y-intercept: $(0, 0)$
• (d)
  • Domain: $(-\infty, \infty)$
  • Range: $(-\infty, 64]$

Would you like more details on any of the steps? Here are some related questions to explore further:

1. How can completing the square help in finding the vertex form of a quadratic function?
2. What is the significance of the vertex in the graph of a quadratic function?
3. How do we determine if a parabola opens upward or downward?
4. Why does every quadratic function have a domain of $(-\infty, \infty)$?
5. Can a quadratic function have more than two x-intercepts? Why or why not?

Tip: To check if a quadratic opens upward or downward, just look at the sign of the coefficient of $x^2$. If it's positive, the parabola opens upward; if it's negative, it opens downward.