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The vertex is a crucial concept when dealing with quadratic functions and parabolas. It is essentially the highest or lowest point on the parabola, depending on the direction it opens. For the function given, \(f(x) = x^2 - 4\), the formula to find the vertex is \[h = \frac{-b}{2a}\] This gives the \(x\)-coordinate of the vertex. In this case, with \(a = 1\), \(b = 0\), and \(c = -4\), we calculate \[h = \frac{-0}{2 \cdot 1} = 0\]. Then, substitute \(h\) back into the function to find the \(y\)-coordinate: \[k = f(0) = 0^2 - 4 = -4\]. Thus, the vertex of this parabola is at the point \((0, -4)\). The vertex helps in determining the symmetry and the shape of the parabola. It is important because it provides a reference point to plot the parabola effectively.

Finding the \(x\)-intercepts of a parabola involves determining the points where the parabola intersects the \(x\)-axis. This means solving for when the function equals zero. For our function \(f(x) = x^2 - 4\), the calculation is straightforward:Set the equation to zero: \[x^2 - 4 = 0\] You factor the quadratic expression: \[(x - 2)(x + 2) = 0\] Each factor gives a root or intercept:

• When \(x - 2 = 0\), \(x = 2\)
• When \(x + 2 = 0\), \(x = -2\)

These solutions, \(x = 2\) and \(x = -2\), indicate that the parabola crosses the \(x\)-axis at the points \((2, 0)\) and \((-2, 0)\). The presence of two intercepts hints at the symmetric nature of the parabola around its vertex.

Sketching a parabola involves using the vertex and \(x\)-intercepts to accurately visualize the curve. With the vertex at \((0, -4)\) and \(x\)-intercepts at \((-2,0)\) and \((2,0)\), you can begin sketching the parabola on a coordinate grid.
Start by plotting these key points. These provide a skeleton to guide your drawing. Since \(a = 1\) for \(x^2\), this positive coefficient indicates the parabola opens upwards.
Parabolas are symmetric about the vertical line through the vertex, known as the axis of symmetry. In this case, it's the \(x\)-axis \(x = 0\).
Draw smoothly connecting the vertex to the \(x\)-intercepts on both sides to form a "U" shape. The symmetry will guide the smooth curve, making sure it peaks at the vertex and passes through the intercepts perfectly.
Sketching by hand, ensures you visualize how each part relates: the vertex's location, the intercepts' placement, and the direction in which the parabola opens. This holistic understanding enriches the grasp of quadratic functions.