Problem 92 - Algebra 2

Question

Determine whether the function has a maximum or minimum value. Then find the value. $$ y=x^{2}+4 x-8 $$

Step-by-Step Solution

Answer

The function has a minimum value at \(y=-4\).

Step 1: Find the derivative of the function

Calculate the derivative of the function \(y = x^{2}+4 x-8\). This is achieved through the power rule, which states that the derivative of \(x^n\) where n is any real number is \(nx^{n-1}\). The derivative of the function, denoted as \(y'\), is therefore \(2x+4\).

Step 2: Set the derivative equal to zero

The maximum or minimum value of a function takes place where its derivative is either zero or undefined. Hence, the derivative \(2x+4=0\) should be solved for x. Subtracting 4 from both sides and then dividing by 2 gives \(x=-2\). This value of x is where the maximum or minimum of the function occurs.

Step 3: Second derivative test

The second derivative test is used to classify the stationary point as a local maximum or minimum. Find the second derivative of the function by taking the derivative of the first derivative from step 1. The derivative of \(2x+4\) is \(2\), which is greater than zero. Therefore, at \(x=-2\), the function has a minimum.

Step 4: Find the minimum value

Substitute \(x=-2\) into the original function \(y = x^{2}+4 x-8\). Therefore, \(y=(-2)^{2}+4(-2)-8=-4\), which is the minimum value of the function.

Key Concepts

Finding Derivatives

To understand quadratic functions, we often start by finding their derivatives. This is crucial to analyzing the behavior of the function. For the quadratic equation \( y = x^2 + 4x - 8 \), we apply the power rule in calculus. The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \). Applying this rule to each term:

• The derivative of \( x^2 \) is \( 2x \).
• The derivative of \( 4x \) is \( 4 \).
• The derivative of \( -8 \) is \( 0 \), since constants become zero.

Combining these, we find the derivative of the given function is \( y' = 2x + 4 \). This derivative helps us identify critical points which could be potential maximums or minimums.

Second Derivative Test

After finding the first derivative, it’s important to determine the nature of the critical points. Here, the second derivative test comes into play. This test helps us to classify whether the function has a local maximum, local minimum, or neither at the critical point. We take the derivative of the first derivative found earlier, so for \( y' = 2x + 4 \), we differentiate once more. This yields the second derivative: \( y'' = 2 \).

• If \( y'' > 0 \), the function has a local minimum at the critical point.
• If \( y'' < 0 \), the function has a local maximum.
• If \( y'' = 0 \), the test is inconclusive.

For this problem, since \( y'' = 2 \) which is positive, the function has a minimum at the critical point \( x = -2 \).

Finding Minimum or Maximum

With quadratic functions, finding a minimum or maximum value requires critical point analysis. We set the first derivative, \( y' = 2x + 4 \), equal to zero: \[ 2x + 4 = 0 \]Solving this, we find \( x = -2 \). This x-value is where our function hits a critical point. Since our second derivative test indicates a minimum, we substitute \( x = -2 \) back into the original function to find the minimum value:\[ y = (-2)^2 + 4(-2) - 8 \]\[ y = 4 - 8 - 8 \]\[ y = -4 \]Thus, the minimum value of the function is \( -4 \) when \( x = -2 \). This underscores that understanding the nature of derivatives helps in pinpointing precise minimum or maximum values.

Critical Points

Critical points in calculus occur where the first derivative of a function is zero or undefined. They are key to identifying where the function reaches extreme values, either maximum or minimum.For our function, \( y = x^2 + 4x - 8 \), setting the derivative \( y' = 2x + 4 \) to zero gives:\[ 2x + 4 = 0 \]Solving for \( x \), we get \( x = -2 \). This is our critical point.

• First, we find potential locations for maximum and minimum values.
• Through further testing, using the second derivative, we classify these points.

The critical point \( x = -2 \) leads us to a minimum value in this scenario. Recognizing these critical points simplifies the process of analyzing and predicting the behavior of quadratic functions.