Problem 129 - Calculus Volume 1

Question

Find the local and absolute minima and maxima for the functions over \((-\infty, \infty)\). \(y=x^{2}+4 x+5\)

Step-by-Step Solution

Answer

The absolute minimum is at \((-2, 1)\). There is no maximum.

Step 1: Identify the Type of Function

The function given is a quadratic function in the form of \( y = ax^2 + bx + c \). This is a parabola, and since \( a = 1 \), which is positive, the parabola opens upwards. This means it will have a vertex, which will act as a minimum point.

Step 2: Find the Vertex of the Parabola

The vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using the vertex formula \( x = -\frac{b}{2a} \). For our function, \( a = 1 \) and \( b = 4 \). Substituting these values, \( x = -\frac{4}{2 \times 1} = -2 \).

Step 3: Calculate the Minimum Value

Substitute \( x = -2 \) back into the original function to find the minimum \( y \)-value. This gives \( y = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1 \). So, the vertex is \((-2, 1)\).

Step 4: Determine Local and Absolute Extrema

Since this is a quadratic function opening upwards and defined over all real numbers, the vertex \((-2, 1)\) is the absolute minimum. There are no local or absolute maxima because the parabola extends indefinitely upwards.

Key Concepts

Understanding Quadratic Functions

Quadratic functions are one of the foundational elements in calculus, known for their characteristic U-shaped graphs called parabolas. A generic quadratic function is written as \( y = ax^2 + bx + c \). Here:

• \(a\), \(b\), and \(c\) are numeric coefficients
• \(a eq 0\) determines the direction in which the parabola opens

If \(a > 0\), the parabola opens upwards, resembling a smile. Conversely, if \(a < 0\), it opens downwards, like a frown.
These functions are pivotal in modeling various real-world problems, such as projectiles' paths and optimization problems. Their simple form makes them convenient for determining crucial points like the vertex and optimizing values.

Locating the Parabola Vertex

The vertex of a parabola is either its highest or lowest point, depending on whether it opens upwards or downwards. For the function \( y = ax^2 + bx + c \), the vertex can be calculated using the formula \( x = -\frac{b}{2a} \). This x-coordinate gives the horizontal position of the vertex on the graph.
Once the x-coordinate is found, substitute it back into the quadratic equation to solve for \( y \), giving you the exact vertex point \((x, y)\).
For example, given \( y = x^2 + 4x + 5 \), you would have:

• \( a = 1 \)
• \( b = 4 \)
• \( x = -\frac{4}{2 \times 1} = -2 \)

Plugging \( -2 \) into the function, you get the y value, yielding the vertex \((-2, 1)\). Because the parabola opens upward, the vertex denotes the lowest point on the graph.

Understanding Extrema in Quadratic Functions

In calculus, the term "extrema" refers to the maximum and minimum values of a function. For a quadratic function that opens upward, like \( y = ax^2 + bx + c \) with \( a > 0 \), the vertex serves as the point of minimum value.
This minimum is called the absolute minimum since it is the smallest y-value the function will take across its entire domain, \(( -\infty, \infty )\). Quadratic functions have no maximum if they open upwards because the y-values can increase infinitely.
Thus, analyzing the example \( y = x^2 + 4x + 5 \), we find that the absolute minimum is at the vertex \((-2, 1)\). This is critical as it tells us where the function reaches its lowest point, offering insights into potential solutions for problems modeled by this quadratic function.