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A parabola is a U-shaped curve that is the graph of a quadratic function. It can open either upward or downward. The direction it opens depends on the coefficient of the squared term in the quadratic function. In the vertex form of a quadratic function, which is given by






(x-h)^2 + k






0\), the parabola opens upward. If \(a < 0\), the parabola opens downward. Understanding these basic properties will help you analyze the graph of any quadratic function effectively.

The x-intercept(s) of a quadratic function are the points where the graph crosses the x-axis. To find the x-intercepts, you need to solve the equation \(k(x) = 0\). For the given function \(k(x) = 2(x-3)^2 - 2\), you set it equal to zero and solve: \[0 = 2(x-3)^2 -2\]
Solving this equation, we get: \[(x-3)^2 = 1\]
Therefore, \[x-3 = \pm 1\], which simplifies to \(x = 4\) or \(x = 2\). So, the x-intercepts are \((4, 0)\) and \((2, 0)\). These points are where the graph touches the x-axis.

The y-intercept of a quadratic function is the point where the graph crosses the y-axis. To find the y-intercept, set \(x = 0\) in the function and solve for \(k(x)\). For \(k(x) = 2(x-3)^2 - 2\): \[k(0) = 2(0-3)^2 - 2\]
This simplifies to: \[k(0) = 2(9) - 2\]
\[k(0) = 18 - 2\]
\[k(0) = 16\]. So, the y-intercept is \((0, 16)\). This is the point where the graph crosses the y-axis, providing important information about the function's behavior.

The axis of symmetry of a parabola is a vertical line that divides the graph into two mirror images. For a quadratic function in vertex form \(k(x) = a(x-h)^2 + k\), the axis of symmetry is given by the equation \(x = h\). In our example, \(k(x) = 2(x-3)^2 - 2\), the axis of symmetry is \(x = 3\). This vertical line passes through the vertex and helps in identifying the mirrored nature of the graph.

The domain and range provide information about the set of possible input and output values of a function: The domain of a quadratic function is all real numbers, written as \((-\br\br\br\br \infty \infty) \$, because a parabola extends indefinitely in both directions along the x-axis.The range,however depends on the direction the parabola opens.Since our function \)I(2x-3)^2 - 2\(with opening upwards'circle '\)a > 0\(, the range starts from the minimum value and extends to \)\infty: [\br -2 positive \infty] \$Therefore,the function's value can never below \(-2\),but can be any value.above it. Understanding the domain and range helps you grasp how the quadratic function behaves across inputs.