Problem 83 - Elementary Algebra

Question

Determine the maximum or minimum \(y\) -value. \(y=x 2-10 x+1\)

Step-by-Step Solution

Answer

The minimum y-value is -24.

Step 1: Identify the Function Type

The given function is a quadratic function, which can be written in the standard form as \( y = ax^2 + bx + c \). In this case, \( a = 1 \), \( b = -10 \), and \( c = 1 \). Quadratic functions form a parabola when graphed.

Step 2: Determine the Parabola Direction

Since the coefficient \( a = 1 \) is positive, the parabola opens upwards. This means the function has a minimum point (vertex) rather than a maximum point.

Step 3: Calculate the Vertex's x-value

Use the vertex formula for the x-coordinate of the vertex: \( x = -\frac{b}{2a} \). Substitute the values: \( x = -\frac{-10}{2 \cdot 1} = 5 \).

Step 4: Find the Minimum y-value

Substitute \( x = 5 \) back into the function to find the minimum \( y \)-value: \( y = (5)^2 - 10(5) + 1 = 25 - 50 + 1 = -24 \). Thus, the minimum \( y \)-value is \(-24\).

Key Concepts

Vertex of a Parabola

The vertex of a parabola is its highest or lowest point, depending on the parabola's direction. This crucial point can be located on the graph of a quadratic function. A quadratic function in the form \( y = ax^2 + bx + c \) has a vertex given by the coordinates \( (x, y) \). To find the x-coordinate, use the formula:

• \( x = -\frac{b}{2a} \)

This formula arises from completing the square or using calculus. For the function \( y = x^2 - 10x + 1 \), we find the vertex's x-coordinate as \( x = 5 \). Substitute this back into the original equation to get the y-coordinate, as we did, getting \( y = -24 \). Thus, the vertex here is at \( (5, -24) \). The vertex can tell us a lot about the parabola, such as whether it's an upward or downward curve.

Minimum and Maximum Value

Quadratic functions can either have a minimum or a maximum value. This occurs at their vertex, and is determined by the coefficient \( a \) in the standard form.

• If \( a > 0 \), the parabola opens upwards, meaning the vertex represents a minimum value.
• If \( a < 0 \), the parabola opens downwards, and the vertex is at the maximum value.

In this exercise, we have \( a = 1 \), which is positive, indicating an upward-opening parabola, so the vertex provides a minimum value. The unique minimum y-value for the function \( y = x^2 - 10x +1 \) is \(-24\), which is achieved when \( x = 5 \). Thus, understanding the minimum or maximum values help in predicting the behavior of the function.

Parabola Direction

The direction in which a parabola opens is crucial for understanding its graph and determining its properties like the vertex.

• When the quadratic coefficient \( a > 0 \), the parabola opens upwards.
• When \( a < 0 \), the parabola opens downwards.

The direction tells us where the extreme value (minimum or maximum) of the function is located. With \( a = 1 \) in our function \( y = x^2 - 10x + 1 \), we have an upwards-opening parabola. This means that the vertex of the parabola is the lowest point on its graph, and not the highest, reflecting the minimum value of \( -24 \). Understanding this aspect ensures accurate graph representations and problem-solving.

Standard Form of a Quadratic Function

The standard form of a quadratic function is \( y = ax^2 + bx + c \). This is pivotal for determining the graph's shape, vertex location, and the parabola's direction.

• The coefficient \( a \) lets us know if the parabola opens upwards or downwards.
• The vertex can be calculated using the formula \( x = -\frac{b}{2a} \).
• By substituting \( x \) back into the equation, the vertex's y-value can be found.

Given the exercise function \( y = x^2 - 10x + 1 \), we see it is already in standard form with \( a = 1 \), \( b = -10 \), and \( c = 1 \). This form is intuitive for solving many problems, like finding the minimum or maximum value, and plotting the function on a graph. Understanding the standard form provides immediate insight into the parabola's general properties.