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The vertex of a quadratic function is a key feature that provides important information about the graph's maximum or minimum point. A quadratic function in standard form is expressed as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. The formula to find the vertex \((h, k)\) is crucial. To determine \( h \), use the vertex formula:

• \( h = -\frac{b}{2a} \)

Let’s look at the function \( f(x) = x^2 + 4x + 4 \). Here, \( a = 1 \), \( b = 4 \), so substituting these into the formula gives:

• \( h = -\frac{4}{2 \times 1} = -2 \)

Once \( h \) is calculated, we find \( k \) by substituting \( h \) back into the original function:

• \( k = (-2)^2 + 4(-2) + 4 = 0 \)

Thus, the vertex of the graph is \((-2, 0)\). This point is very important because it tells us that the graph of the function is centered at this point and because the parabola opens upwards (since \( a > 0 \)), it is a minimum point.

Identifying the \( x \)-intercepts of a quadratic function involves finding the points where the graph crosses the \( x \)-axis. These occur where the value of the function \( f(x) \) is zero. For \( f(x) = x^2 + 4x + 4 \), we set the function equal to zero to find these intercepts:

• \( x^2 + 4x + 4 = 0 \)

To solve this, we look for factors of the equation. It can be factored as:

• \( (x + 2)^2 = 0 \)

This indicates that there is a repeated root \( x = -2 \). Consequently, the function has one \( x \)-intercept at \((-2, 0)\). This is the same point as the vertex, meaning the parabola just touches the \( x \)-axis at this point without crossing it. This makes \((-2, 0)\) both the vertex and the sole \( x \)-intercept of the quadratic function.

The \( y \)-intercept of a quadratic function is the point where the graph crosses the \( y \)-axis. This is easily found by substituting \( x = 0 \) into the function, as it represents the point on the \( y \)-axis. For the function \( f(x) = x^2 + 4x + 4 \), by substituting \( x = 0 \), we identify the \( y \)-intercept:

• \( f(0) = (0)^2 + 4(0) + 4 = 4 \)

This gives the \( y \)-intercept as \((0, 4)\). This point shows where the graph of the function crosses the \( y \)-axis. The parabola extends upward from this point. In essence, the \( y \)-intercept provides initial insight into how the parabola stretches along the \( x \)-axis.