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The vertex of a quadratic function is a key point on its graph, often representing the minimum or maximum value of the function depending on the direction of the parabola. The formula used to find the x-coordinate of the vertex is given by \[ x = -\frac{b}{2a} \]where \(a\) and \(b\) are coefficients from the standard quadratic equation \(ax^2 + bx + c\). For the quadratic function in the exercise, with coefficients \(a = \frac{1}{2}\) and \(b = 4\), the x-coordinate of the vertex is calculated as \(-4\). To find the y-coordinate, substitute \(x = -4\) back into the quadratic function. This results in:\[ f(-4) = \frac{1}{2}(-4)^2 + 4(-4) + \frac{15}{2} = \frac{-9}{2} \]Thus, the vertex of the parabola is \((-4, -\frac{9}{2})\). The vertex provides important information about the graph's shape, acting as the 'turning point' of the parabola. Understanding the vertex helps with sketching and analyzing the function's properties.

• Vertex formula: \(x = -\frac{b}{2a}\)
• X-coordinate: \(-4\)
• Y-coordinate after substitution: \(-\frac{9}{2}\)

The x-intercepts of a quadratic function are the points where the graph crosses the x-axis. To find these intercepts, set the quadratic function equal to zero and solve for \(x\). This represents the roots of the equation:\[ \frac{1}{2}x^2 + 4x + \frac{15}{2} = 0 \]Multiply through by 2 to clear the fraction, yielding:\[ x^2 + 8x + 15 = 0 \]Factor this quadratic equation to find:\[(x + 3)(x + 5) = 0 \]Thus, the x-intercepts are \(x = -3\) and \(x = -5\), corresponding to the points \((-3, 0)\) and \((-5, 0)\) on the graph. These intercepts are crucial as they assist in understanding how the function behaves at different values of \(x\). They provide insight into the function's real-world applications, such as moments when a projectile hits the ground.

• Set function to zero: \(f(x) = 0\)
• Solve: Roots at \(x = -3\) and \(x = -5\)
• X-intercept points: \((-3, 0)\), \((-5, 0)\)

The y-intercept is the point where the graph of the quadratic function crosses the y-axis. This happens when \(x\) is zero. Consequently, to find the y-intercept, simply evaluate the function at \(x = 0\):\[ f(0) = \frac{1}{2}(0)^2 + 4(0) + \frac{15}{2} \]This simplifies to:\[ f(0) = \frac{15}{2} \]Hence, the y-intercept is at the point \(\left(0, \frac{15}{2}\right)\). Knowing the y-intercept is beneficial as it provides a specific starting point on the graph. This can help in sketching the function and understanding its initial value when graphing by hand.

• Set \(x = 0\)
• Evaluate: \(f(0) = \frac{15}{2}\)
• Y-intercept point: \(\left(0, \frac{15}{2}\right)\)

The direction in which a parabola opens is dictated by the sign of the leading coefficient \(a\) in the quadratic equation. This provides significant insight into the behavior of the graph under analysis.

• If \(a > 0\), the parabola opens upwards, resembling a U-shape.
• If \(a < 0\), the parabola opens downwards, forming an upside-down U.

In the original exercise, the value of \(a\) is \(\frac{1}{2}\), which is positive. Therefore, the graph of the quadratic function opens upwards. This is useful when sketching the parabola, as it shows the general shape and direction, helping to predict the function's behavior over its domain. Knowing the direction is also vital when determining the vertex as a minimum or maximum point.

• Positive \(a\) indicates upwards opening
• Negative \(a\) indicates downwards opening
• \(a = \frac{1}{2}\): Upwards parabola

In a quadratic equation of the form \(ax^2 + bx + c\), the values \(a\), \(b\), and \(c\) are known as the coefficients. Each coefficient plays a unique role:

• \(a\): The coefficient of \(x^2\) determines the parabola's direction and width. Larger \(|a|\) values make the parabola narrower.
• \(b\): The coefficient of \(x\) affects the symmetry and location of the vertex along the x-axis.
• \(c\): The constant term represents the y-intercept of the function.

For the given equation \(f(x) = \frac{1}{2}x^2 + 4x + \frac{15}{2}\):

• \(a = \frac{1}{2}\), affecting the graph's opening (upwards, due to \(a > 0\)) and width.
• \(b = 4\), influencing the horizontal position of the vertex.
• \(c = \frac{15}{2}\), setting the y-intercept at \(\left(0, \frac{15}{2}\right)\).

Understanding these coefficients is crucial for interpreting how each part of the quadratic function contributes to the graph's overall shape and position. This knowledge is imperative when graphing or applying quadratic functions in various scenarios.