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In a quadratic equation of the form \( y = ax^2 + bx + c \), the vertex is a crucial point. It represents the peak or the trough of the parabola, depending on the sign of the coefficient \( a \). To locate the vertex, use the formula \( x = -\frac{b}{2a} \). Here, the values \( a \) and \( b \) are from our specific quadratic equation. In the given problem, \( a = -4 \) and \( b = 12 \). Substituting these values into the vertex formula, we get: \( x = -\frac{12}{2(-4)} = 1.5 \).This means the vertex is located at \( x = 1.5 \). Next, by plugging this value back into the quadratic equation, we can find the y-coordinate of the vertex. Together, these coordinates (\(1.5, 4\)) are the vertex of the parabola.

In the context of a quadratic function, the maximum or minimum value corresponds to the y-coordinate of the vertex. Since the coefficient \( a \) is negative in our equation, \( y = -4x^2 + 12x - 5 \), the parabola opens downward, indicating a maximum value.To find this maximum value, we already calculated the x-coordinate of the vertex, \( x = 1.5 \). Now, by substituting \( x = 1.5 \) back into the quadratic equation, we found \( y = 4 \). Thus, the maximum value of the given quadratic function is \( 4 \). This is the highest point on the graph of the parabola.

A parabola is the graph of a quadratic function, and it has a distinct U-shape. Depending on the sign of the coefficient \( a \), the parabola either opens upwards or downwards. For positive \( a \), it opens upwards, creating a minimum value at the vertex. For negative \( a \), it opens downwards, producing a maximum value at the vertex.
In our problem, with \( a = -4 \), the parabola opens downwards.
Visualizing this, the vertex (\(1.5, 4\)) lies at the highest point because the arms of the parabola extend downward. Understanding the nature of parabolas helps in determining whether a quadratic equation will have a maximum or minimum value, and how the graph behaves as \( x \) moves away from the vertex.