Problem 4 - A Graphical Approach to College Algebra

Question

Solve each problem. Do not use a calculator. Find the minimum \(y\) -value on the graph of \(y=5 x^{2}+30 x+17\)

Step-by-Step Solution

Answer

The minimum \( y \)-value of the function is \(-28\).

Step 1: Identify the Type of Function

The given function is quadratic: \( y = 5x^2 + 30x + 17 \). Quadratic functions form a parabola when graphed.

Step 2: Determine the Vertex Form

The vertex of a parabola given by the equation \( y = ax^2 + bx + c \) has the \( x \)-coordinate at \( x = -\frac{b}{2a} \). Plugging in the values, we get \( x = -\frac{30}{2 \times 5} = -3 \).

Step 3: Calculate the Minimum y-value

Substitute \( x = -3 \) back into the original equation to find \( y \): \(y = 5(-3)^2 + 30(-3) + 17 = 45 - 90 + 17 = -28 \).

Step 4: Interpret the Result

Since the parabola opens upwards (as indicated by the positive \( a \) value, \( a = 5 \)), the vertex represents the minimum point of the function. Thus, the minimum \( y \)-value is \(-28\).

Key Concepts

Parabola

A parabola is a U-shaped curve that is the graph of a quadratic function. Quadratic functions generally have the standard form of \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. The shape and direction of a parabola depend on the value of \( a \).
If \( a \) is positive, the parabola opens upwards, resembling the shape of a smile. If \( a \) is negative, it opens downwards, like a frown. This direction is important for determining whether you will find a minimum or maximum point.
In the problem at hand, the given function produces a parabola opening upwards since \( a = 5 \), which is positive. This is why the vertex, where the minimum or maximum value exists, represents a minimum \( y \)-value.

Vertex Form

The vertex form of a parabola helps in easily identifying the vertex or the peak/valley of the parabola. The standard vertex form equation is \( y = a(x - h)^2 + k \), where \( (h, k) \) represents the vertex of the parabola.
However, when dealing with the standard quadratic equation \( y = ax^2 + bx + c \), you can convert it to the vertex form by completing the square, or you can find the vertex directly.
To find the x-coordinate of the vertex directly from the standard form, use the formula \( x = -\frac{b}{2a} \). In this function, substituting the values from the problem, \( x = -\frac{30}{2 \times 5} = -3 \). Next, substitute \( x = -3 \) back into the quadratic function to calculate the corresponding \( y \)-value, making \( (h, k) = (-3, -28) \). This \( y \)-value is crucial for verifying the minimum point.

Minimum y-value

In any quadratic function that forms a parabola opening upwards, the vertex represents the lowest point on the graph, also known as the minimum \( y \)-value. Thus, the task of finding the minimum \( y \)-value requires identifying the vertex.
Starting with finding the x-coordinate of the vertex using the formula \( x = -\frac{b}{2a} \), substitute it back into the function to solve for \( y \).
In the example problem, once the \( x \)-coordinate \( x = -3 \) was found, it was plugged back into the quadratic equation \( y = 5x^2 + 30x + 17 \). This calculation yields \( y = -28 \), which is the minimum \( y \)-value.

• This minimum value is confirmed because the parabola opens upwards.
• It effectively represents the smallest output value the function can achieve as \( x \) varies.

Understanding this point can help interpret the behavior and constraints of functions in practical applications.