91Ó°ÊÓ

The vertex of a parabola is a key component, representing the highest or lowest point, commonly referred to as the "turning point". It provides the parabola's maximum or minimum value.
To determine the vertex of a quadratic function like \(f(x) = x^2 - 4x + 4\), we use the vertex formula:

• \(x_v = \frac{-b}{2a}\)

Here, \(a\) is the coefficient of \(x^2\) and \(b\) is the coefficient of \(x\). For \(f(x) = x^2 - 4x + 4\), \(a = 1\) and \(b = -4\), giving us:

• \(x_v = \frac{-(-4)}{2 \times 1} = 2\)

Once you've calculated \(x_v\), substitute it back into the function to find \(y_v\), the y-coordinate of the vertex:

• \(y_v = f(2) = (2)^2 - 4(2) + 4 = 0\)

Therefore, the vertex is at \((2, 0)\). The vertex also serves as the axis of symmetry of the parabola, with the equation \(x = 2\). This divides the parabola into two matching halves.

The x-intercepts of a quadratic function are the points where the graph meets the x-axis. At these points, the function value is zero, so to find them, we solve \(f(x) = 0\).
For \(f(x) = x^2 - 4x + 4\), we solve:

• \(x^2 - 4x + 4 = 0\)

This is a perfect square trinomial and can be factored as:

• \((x - 2)^2 = 0\)

Solving for \(x\) gives:

• \(x - 2 = 0 \Rightarrow x = 2\)

Thus, there is only one x-intercept at \((2, 0)\). It's important to notice that when a quadratic function has only one x-intercept, the parabola touches the x-axis, often indicating that the vertex itself lies on the x-axis.

Sketching a parabola involves identifying its vertex, intercepts, and direction of opening. These elements help visualize the graph's basic shape.
For \(f(x) = x^2 - 4x + 4\):

• The vertex is \((2, 0)\).
• There's a single x-intercept at \((2, 0)\).

Since \(a = 1\) is positive, the parabola opens upwards. Consider these steps for sketching:

• Plot the vertex \((2, 0)\) on the coordinate plane.
• Mark the x-intercept at \((2, 0)\); this is also the vertex.
• The axis of symmetry, \(x = 2\), helps visualize the parabola's symmetry.

With these, draw a symmetrical curve opening upwards from the vertex. Remember that the parabola is steepest near its vertex, gradually flattening out as it extends upward. As there is only one intercept, the parabola just "touches" the x-axis at its vertex.