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A quadratic function is mathematically represented as \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants with \( a \) not equal to zero. This type of function produces a parabolic graph—the shape of a U or an upside-down U depending on the value of \( a \). If \( a > 0 \), the parabola opens upwards, and if \( a < 0 \), it opens downwards.

When graphing a quadratic function like \( y = x^2 - 2x - 1 \), we first note the coefficients, which give us clues about the function's key characteristics such as its width, direction, and y-intercept. For our function, the \( a \) coefficient is 1, meaning the parabola will be of standard width and open upwards. The \( b \) coefficient is -2, and \( c \) is -1, which affects the position of the parabola on the Cartesian plane. In real-life applications, quadratic functions can model various phenomena, including projectile motion and profit optimization.

The vertex of a parabola is a significant feature acting as the peak or the lowest point depending on whether the parabola opens downwards or upwards, respectively. For the given quadratic function \( y = x^2 - 2x - 1 \), we've determined the vertex to be at the point \( (1, -2) \).

Finding the vertex involves using the formulas \( h = -\frac{b}{2a} \) and \( k = c - \frac{b^2}{4a} \), derived from completing the square process of the quadratic equation. In our example, after computation, we found that \( h = 1 \) and \( k = -2 \), signaling that our parabola's vertex lies at \( (1, -2) \). The graph's vertex is crucial for sketching the parabola accurately as it provides a starting point from which the curve extends.

X-intercepts are the points where the graph of the quadratic function crosses the x-axis. These are the values of \( x \) for which \( y \) is zero. To find the x-intercepts of the quadratic function \( y = x^2 - 2x - 1 \), you set the function equal to zero and solve for \( x \).

The discriminant, represented by \( b^2 - 4ac \), tells us the nature of the roots or x-intercepts. A positive discriminant, as we have with the value 8 in our exercise, indicates two real and distinct intercepts. In our step by step example, the x-intercepts were found to be at the points \( 1 + \sqrt{2} \) and \( 1 - \sqrt{2} \), which are crucial in sketching the graph as they define where the parabola meets the x-axis. Without x-intercepts, one might only presume the parabola's path based on its vertex and direction.

The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), is a universal solution for the roots of any quadratic equation of the form \( ax^2 + bx + c = 0 \). This powerful formula takes into account the coefficients of the equation and provides the exact values of the x-intercepts when substituted correctly.

In our exercise, applying the quadratic formula allowed us to find the precise x-intercepts \( 1 + \sqrt{2} \) and \( 1 - \sqrt{2} \). Even when the quadratic cannot be factored easily, the quadratic formula is a reliable method for locating where the function will intersect the x-axis. It's essential to correctly substitute the values of \( a \), \( b \), and \( c \) into the formula and carefully perform the arithmetic to ensure accuracy in the resulting intercepts.