Problem 53 - Algebra for College Students

Question

Find the maximum or minimum value of \(y\) for each function. $$y=x^{2}-8$$

Step-by-Step Solution

Answer

The minimum value of \(y\) is \(-8\).

Step 1: Identify the Type of Function

Recognize that the function given is a quadratic function of the form \(y = ax^2 + bx + c\). In this case, the function is \(y = x^2 - 8\).

Step 2: Determine the Leading Coefficient

Observe the coefficient of the \(x^2\) term, which is \(1\). Since \(1 > 0\), this means the parabola opens upwards.

Step 3: Identify the Vertex

Quadratic functions of the form \(y = ax^2 + bx + c\) have their vertex at the point \((h, k)\). For \(y = x^2 - 8\), the vertex is at \( (0, -8) \). Because the parabola opens upwards, this vertex represents the minimum value of the function.

Step 4: Find the Minimum Value

The minimum value of \(y\) occurs at the vertex. Substituting \(x = 0\) into the function, \(y = (0)^2 - 8 = -8\). Therefore, the minimum value of \(y\) is \(-8\).

Key Concepts

parabola

A quadratic function is often represented as a parabola in a coordinate plane. This type of curve is symmetrical and can either open upwards or downwards depending on the coefficient of the squared term. In our example, the quadratic function is given by \(y = x^{2} - 8\).

For any quadratic function in the standard form \(y = ax^{2} + bx + c\), if the coefficient \(a\) is positive, the parabola opens upwards. If \(a\) is negative, the parabola opens downwards.

In our problem, \(a = 1\), which is positive, so the parabola opens upwards.

vertex

The vertex of a quadratic function is a critical point that represents either the maximum or the minimum value of the function. The formula for finding the vertex of a quadratic function \(y = ax^{2} + bx + c\) is \((h, k)\) where \(h = -\frac{b}{2a}\) and \(k\) is the value of the function at \(h\).

For the given function \(y = x^{2} - 8\), since there is no linear term (i.e., \(b = 0\)), the vertex occurs at \(h = 0\). Substituting \(x = 0\) into the equation, we get \(y = (0)^2 - 8 = -8\). Hence, the vertex of the function is at the point \(0, -8\).

minimum value

The minimum or maximum value of a quadratic function occurs at its vertex. For the given function \(y = x^2 - 8\), we have already identified that the vertex is at \( (0, -8) \).

Since the parabola opens upwards (as established by the positive coefficient of \(x^2\)), the vertex represents the minimum value of the function. Therefore, the minimum value of \(y\) for the given quadratic function is \-8\.

To summarize:

• A parabola that opens upwards has a minimum value at the vertex.


• The minimum value is the y-coordinate of the vertex point.


• For \(y = x^{2} - 8\), the minimum value is -8.