The function provided is:

$f(x) = -x^2 + 4x + 5$

This is a quadratic function. To analyze this function, let's go over some key features:

1. Vertex Form (Max/Min Point)

Since the coefficient of $x^2$ is negative, the parabola opens downwards, meaning it has a maximum point. We can find the vertex (maximum) by completing the square or using the formula for the vertex:

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

Here, $a = -1$, $b = 4$. Substituting these into the vertex formula:

$x = -\frac{4}{2(-1)} = 2$

Now, substitute $x = 2$ back into the original function to find $f(2)$:

$f(2) = -(2)^2 + 4(2) + 5 = -4 + 8 + 5 = 9$

So, the vertex is at $(2, 9)$, which is the maximum point of the parabola.

2. Axis of Symmetry

The axis of symmetry is the vertical line that passes through the vertex. Its equation is:

$x = 2$

3. Y-Intercept

The y-intercept occurs when $x = 0$:

$f(0) = -(0)^2 + 4(0) + 5 = 5$

So, the y-intercept is at $(0, 5)$.

4. X-Intercepts (Roots)

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

$-x^2 + 4x + 5 = 0$

Multiply through by -1 for simplicity:

$x^2 - 4x - 5 = 0$

Now, factor the quadratic:

$(x - 5)(x + 1) = 0$

Thus, the solutions are:

$x = 5 \quad \text{and} \quad x = -1$

The x-intercepts are at $(5, 0)$ and $(-1, 0)$.

Summary of Key Points:

• Vertex: $(2, 9)$
• Axis of symmetry: $x = 2$
• Y-intercept: $(0, 5)$
• X-intercepts: $(5, 0)$ and $(-1, 0)$

Let me know if you want more details or have any questions!

Here are some related questions for further exploration:

1. How do you find the vertex of any quadratic function in general form?
2. How do transformations affect the graph of a quadratic function?
3. How would the graph change if the leading coefficient was positive?
4. How do you use the quadratic formula to solve quadratic equations?
5. What are the practical applications of finding the vertex of a parabola?

Tip: A negative coefficient for $x^2$ means the parabola opens downward, indicating a maximum point.