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Quadratic functions, like the one provided in the exercise, are often characterized by their parabolic graphs. Finding the vertex is essential as it indicates the turning point of the parabola. To find the vertex of a quadratic function written in the form \( ax^2 + bx + c \), use the vertex formula, which is a handy tool:

• Formula: \( x = -\frac{b}{2a} \).

For example, in the equation \( f(x) = x^2 - 2x + 2 \), you plug in \( a = 1 \) and \( b = -2 \) into the formula, giving \( x = 1 \). Once you have the x-coordinate of the vertex, substitute it back into the function to find the y-coordinate. Therefore, the vertex in this case is \((1, 1)\). This point is crucial for sketching and analyzing the graph of the quadratic function.

Graphing a parabola involves several steps and facts. The graph of a quadratic function is always a parabola. Here's what you should keep in mind:

• Direction: The parabola opens upwards if \( a > 0 \) and downwards if \( a < 0 \).
• Vertex: The vertex is either the minimum or maximum point, depending on the direction of the parabola.
• Line of Symmetry: The vertical line that passes through the vertex, given by \( x = \text{vertex}\_x \).

In our function, \( f(x) = x^2 - 2x + 2 \), since \( a = 1 \) is positive, the parabola opens upwards. The graph is symmetric around the line \( x = 1 \). Knowing the vertex \((1, 1)\) and the y-intercept \((0, 2)\), you can sketch the parabola as a U-shaped curve centered around the vertex.

X-intercepts, also known as roots or zeros of a function, are the points where the graph crosses the x-axis. To find x-intercepts, set the quadratic function to zero: \( f(x) = 0 \), and solve for \( x \). This requires solving the equation \( ax^2 + bx + c = 0 \). You can use the quadratic formula:

• Formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In our example, \( f(x) = x^2 - 2x + 2 \), this formula results in a negative discriminant \( b^2 - 4ac = -4 \). Since the discriminant is negative, this quadratic function doesn't have any real x-intercepts. Thus, the parabola doesn’t touch the x-axis, confirming no real solutions.

The y-intercept of a quadratic function is where the graph intersects the y-axis. It is easy to find by evaluating the function at \( x = 0 \). This gives the constant term \( c \) in the quadratic equation \( f(x) = ax^2 + bx + c \).For example, in the given function \( f(x) = x^2 - 2x + 2 \), setting \( x = 0 \) yields \( f(0) = 2 \). This means the y-intercept is at the point \((0, 2)\). The y-intercept provides a crucial anchor point for sketching the graph and helps understand how the parabola behaves on the graph around the vertical axis.