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In quadratic functions, the standard form is a way to gracefully express the function to understand its properties better. The standard form of a quadratic function is written as \( f(x) = a(x-h)^2 + k \), where \( (h, k) \) represents the vertex of the parabola.

To express a quadratic function in standard form, like the given \( f(x) = x^2 + 8x \), we use a method called completing the square. This process involves reshaping the quadratic into a perfect square trinomial, which helps us reframe the function neatly. Here's a step-by-step breakdown:

• Identify the coefficient of \( x \), which is 8.
• Divide this coefficient by 2, obtaining 4.
• Square this result to get 16.
• Add and subtract 16 within the function: \( f(x) = (x^2 + 8x + 16) - 16 = (x+4)^2 - 16 \).

This transformation results in the standard form, \( f(x) = (x+4)^2 - 16 \), making it easier to identify the vertex and other characteristics of the function.

The vertex of a quadratic function is a pivotal point described by the coordinates \( (h, k) \) from its standard form, \( f(x) = a(x-h)^2 + k \). It represents the peak or valley of the parabola, depending on whether it opens up or downwards.

From the standard form \( f(x) = (x+4)^2 - 16 \) we derived earlier, the vertex is easily identifiable. Here:

• \( h = -4 \) (since the function is \( (x+4) \)),
• \( k = -16 \).

Therefore, the vertex is located at \((-4, -16)\). This vertex shows the lowest point on the graph of the function because the parabola opens upwards, indicated by the positive coefficient in the square term.

Intercepts are the points where the graph of a function crosses the axes. For a quadratic function, the x-intercepts (roots) and the y-intercept are essential for sketching and understanding the function's behavior.

**X-Intercepts**: To find where the function crosses the x-axis, set \( f(x) = 0 \) and solve for \( x \):

• Starting equation: \( (x+4)^2 - 16 = 0 \).
• Solve: \( (x+4)^2 = 16 \).
• Take the square root: \( x+4 = \pm 4 \).
• Solutions: \( x = 0 \) and \( x = -8 \).

Thus, the x-intercepts are \((0, 0)\) and \((-8, 0)\).

**Y-Intercept**: To find where the function crosses the y-axis, evaluate \( f(0) \) by substituting \( x = 0 \) into the function:

• \( f(0) = 0^2 + 8 \cdot 0 = 0 \).

Therefore, the y-intercept is \((0, 0)\), which coincides with one of the x-intercepts.

Understanding the domain and range of a quadratic function is crucial, as it tells us what values \( x \) can take, and the possible outputs \( f(x) \) can achieve.

**Domain**: Quadratic functions, like any polynomial function, have a domain of all real numbers. This means \( x \) can be any value from \(-\infty\) to \(\infty\).

**Range**: The range of the function is determined by its vertex and the direction in which the parabola opens. Since the function opens upwards and has a vertex at \((-4, -16)\), the range is all real numbers greater than or equal to \(-16\).

Thus, the range is \([-16, \infty)\). This helps us understand that the minimum value is \(-16\), and the function can extend upwards without bound.