Problem 1 - Algebra and Trigonometry

Question

\(1-4\) . The graph of a quadratic function \(f\) is given. (a) Find the coordinates of the vertex. (b) Find the maximum or minimum value of \(f\) $$ f(x)=-x^{2}+6 x-5 $$

Step-by-Step Solution

Answer

The vertex is at (3, 4), and the maximum value of \( f \) is 4.

Step 1: Identify the quadratic function

The quadratic function given is \( f(x) = -x^2 + 6x - 5 \). This is a standard quadratic equation in the form \( ax^2 + bx + c \) where \( a = -1 \), \( b = 6 \), and \( c = -5 \).

Step 2: Use the vertex formula

The vertex of a quadratic function \( ax^2 + bx + c \) is given by the formula \( x = -\frac{b}{2a} \). For the function \( f(x) = -x^2 + 6x - 5 \), substitute \( a = -1 \) and \( b = 6 \) into the formula: \[ x = -\frac{6}{2(-1)} = 3 \]

Step 3: Find the y-coordinate of the vertex

Substitute \( x = 3 \) back into the function \( f(x) = -x^2 + 6x - 5 \) to find the y-coordinate.\[ f(3) = -(3)^2 + 6(3) - 5 = -9 + 18 - 5 = 4 \] Thus, the vertex has coordinates \((3, 4)\).

Step 4: Determine the maximum or minimum value

Since the coefficient of \( x^2 \) in \( f(x) = -x^2 + 6x - 5 \) is negative (\( a = -1 \)), the parabola opens downward. Therefore, the vertex represents the maximum point of the function. The maximum value of \( f \) is \( 4 \) at the vertex.

Key Concepts

Vertex

The vertex of a quadratic function is a significant point that often provides useful insights about the parabola. For the given function, the vertex can be thought of as the peak or the lowest point on the graph, depending on the direction in which the parabola opens. The formula to find the vertex of a quadratic function written in standard form, \( ax^2 + bx + c \), is to compute \( x_v = -\frac{b}{2a} \).
For our function \( f(x) = -x^2 + 6x - 5 \), we already calculated this to be \( x_v = 3 \). To find the complete vertex coordinates, we also need the \( y \)-coordinate. To determine \( y \), substitute \( x_v = 3 \) back into the function:\[ f(3) = -(3)^2 + 6(3) - 5 = 4 \]This gives a vertex at \((3, 4)\). The vertex is crucial because it helps in sketching the graph, understanding the function's behavior, and determining the maximum or minimum values.

Maximum or Minimum Value

The vertex not only helps locate a specific point on the graph, but it also determines whether the quadratic function reaches a peak or a trough. This concept is referred to as the maximum or minimum value of the function.

• If the parabola opens upward (\(a > 0\)), the vertex is the function's minimum.
• If the parabola opens downward (\(a < 0\)), the vertex represents the maximum value.

In the function \(f(x) = -x^2 + 6x - 5\), since the coefficient \(a\) is \(-1\), the parabola opens downward. Therefore, the vertex at \((3, 4)\) gives us the maximum value of the function: 4. This value represents the highest point the parabola will reach on the graph.

Parabola Opening Direction

The direction in which the parabola opens is an important characteristic of quadratic functions. It's determined by the leading coefficient, \(a\), in the quadratic expression \(ax^2 + bx + c\). To understand how this works:

• If \(a > 0\), the parabola opens upward, forming a 'U' shape.
• If \(a < 0\), the parabola opens downward, forming an upside-down 'U'.

For our function \(f(x) = -x^2 + 6x - 5\), the leading coefficient \(a\) is \(-1\). Thus, the parabola opens downward. This affects the vertex, making it a maximum point of the parabola, perfectly situated at the top. The opening direction is vital because it influences the parabola's range and potential extremum points.