91Ó°ÊÓ

When dealing with quadratic functions like this one, identifying the domain and range is crucial. For the function \( f(x) = x - x^2 \), it's important to note that it's a polynomial function. Polynomial functions always have a domain of all real numbers, which means there's no restriction on the values \( x \) can take. This gives us the domain: \((-abla, abla)\).

Finding the range requires a bit more work. We start by locating the vertex, as it provides the maximum or minimum value. In our negative coefficient situation, the graph of \( f(x) = x - x^2 \) opens downward, indicating the presence of a maximum value. The vertex's x-coordinate comes from \( x = -\frac{b}{2a} \). Substituting \( a = -1 \) and \( b = 1 \), we calculate \( x = \frac{1}{2} \). To get the y-value of the vertex, substitute back into \( f(x) \):

\[ f\left(\frac{1}{2}\right) = \frac{1}{2} - \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

Therefore, the vertex is \( \left(\frac{1}{2}, \frac{1}{4}\right) \), and the range is \((-abla, \frac{1}{4}]\), showing all values less than or equal to this maximum.

Intercepts are the points where the graph intersects the axes, showing where the function is zero or where it crosses the y-axis.

**X-Intercepts**
To find the x-intercepts, we set \( f(x) = 0 \) and solve. For \( f(x) = x - x^2 \), factoring gives:

• \( x(x - 1) = 0 \)
• This gives solutions \( x = 0 \) and \( x = 1 \)

Here, the graph intersects the x-axis at points (0,0) and (1,0).

**Y-Intercept**
Finding the y-intercept is straightforward: Set \( x = 0 \) in the function:

\[ f(0) = 0 - 0^2 = 0 \]

The y-intercept is at the origin, (0,0). Thus, this quadratic graph passes through the origin.

For quadratic functions, it's typical to check symmetry by seeing if the function is even or odd. Symmetry makes graphs easier to interpret and plot.

Substitute \( -x \) into \( f(x) \) and compare:

\[ f(-x) = -x - (-x)^2 = -(x - x^2) \]

In this function, \( f(-x) eq f(x) \) and \( f(-x) eq -f(x) \), indicating no symmetry. This informs us that the graph has no specific predictable pattern around the y-axis or the origin.

Identifying extrema, or the highest and lowest points on the graph, is essential for understanding the behavior of a quadratic function.

For \( f(x) = x - x^2 \), the vertex provides the maximum value. The parabola opens downward due to the negative leading coefficient, meaning only a maximum exists. We've already calculated the vertex: \( \left(\frac{1}{2}, \frac{1}{4}\right) \).

Hence, the extremum is a maximum at \( y = \frac{1}{4} \) when \( x = \frac{1}{2} \). This feature helps to sketch the parabola and analyze its range.

Plotting a quadratic function like \( f(x) = x - x^2 \) involves situating critical points and following the parabolic curve's path.

Start by marking your intercepts and vertex:

• X-intercepts: (0,0) and (1,0)
• Y-intercept: (0,0)
• Vertex: \( \left(\frac{1}{2}, \frac{1}{4}\right) \)

Once positioned, these points allow the sketching of a parabola. Since it opens downwards, connect the points smoothly in a downward arc.

This structured approach guarantees an accurate depiction, showcasing the relationship between domain, range, and extrema. The graph is an essential visual aid for understanding quadratic behavior.