Problem 200 - Elementary Algebra

Question

In the following exercises, find the maximum or minimum value. $$ y=-x^{2}+4 x-5 $$

Step-by-Step Solution

Answer

The maximum value is -1 at x = 2.

Step 1 - Identify the coefficients

The given quadratic equation is in the form of \( y = ax^2 + bx + c \). Identify the coefficients: \( a = -1 \), \( b = 4 \), and \( c = -5 \).

Step 2 - Determine the vertex

The vertex form of a quadratic equation gives the maximum or minimum point. For \( y = ax^2 + bx + c \), the x-coordinate of the vertex is found using \( x = \frac{-b}{2a} \). Substitute the given values: \( x = \frac{-4}{2(-1)} = 2 \).

Step 3 - Calculate the y-coordinate of the vertex

Substitute \( x = 2 \) back into the original equation to find the y-coordinate: \( y = - (2)^2 + 4(2) - 5 \). This simplifies to \( y = -4 + 8 - 5 = -1 \).

Step 4 - Determine if it's a maximum or minimum

Since the coefficient \( a = -1 \) is negative, the parabola opens downward, indicating that the vertex represents a maximum value.

Key Concepts

vertex of a parabola

The vertex of a parabola is a crucial point. For any quadratic equation in the standard form \(y = ax^2 + bx + c\), the x-coordinate of the vertex can be found using the formula \(x = \frac{-b}{2a}\). In our exercise, the equation is \(y = -x^2 + 4x - 5\). Here, the coefficients are: \(a = -1\) and \(b = 4\). Plugging these into the formula gives us \(x = \frac{-4}{2(-1)} = 2\).
To find the y-coordinate of the vertex, substitute \(x = 2\) back into the original equation: \(y = -(2)^2 + 4(2) - 5 = -4 + 8 - 5\), which simplifies to \(y = -1\). Therefore, the vertex of the parabola is at the point \((2, -1)\).

maximum and minimum values

The maximum or minimum value of a quadratic equation depends on the direction of the parabola. For \(y = ax^2 + bx + c\), the parabola opens upwards if \(a > 0\) and downwards if \(a < 0\).
In our exercise, since \(a = -1\) (which is less than 0), the parabola opens downwards. This means the vertex represents the maximum point of the parabola.
Therefore, given the vertex \((2, -1)\), the maximum value of the equation \(y = -x^2 + 4x - 5\) is \(y = -1\).

coefficient identification

Identifying the coefficients in a quadratic equation is the first step towards finding the vertex and determining the maximum or minimum values.
Quadratic equations are typically in the form \(y = ax^2 + bx + c\). Here, \(a\), \(b\), and \(c\) represent the coefficients. These coefficients control the shape and position of the parabola.
In our exercise, the quadratic equation is \(y = -x^2 + 4x - 5\). By comparing this with the standard form, we identify:

• \(a = -1\) (which influences the direction and width),
• \(b = 4\) (which affects the slope of the parabola's arms),
• \(c = -5\) (which represents the y-intercept).

parabola direction

The direction a parabola opens is determined by the coefficient \(a\) in the quadratic equation \(y = ax^2 + bx + c\).

• If \(a > 0\), the parabola opens upwards, making the vertex a minimum point.
• If \(a < 0\), the parabola opens downwards, making the vertex a maximum point.

In our exercise, \(a = -1\). Since \(-1 < 0\), the parabola opens downwards. This orientation tells us that the vertex represents the highest or maximum value of the parabola.
Another way to visually confirm this is by plotting the equation's graph, noticing how the arms of the parabola extend downwards, ensuring the vertex at \((2, -1)\) is indeed a peak or maximum.