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A quadratic function is a polynomial function that can be expressed in the form \( y = ax^2 + bx + c \). The graph of a quadratic function is a parabola. Understanding the shape and position of this parabola is crucial for solving optimization problems involving quadratic functions. A positive \( a \) makes the parabola open upwards, like a cup, whereas a negative \( a \) makes it open downwards, like an arch.
In our exercise, the function given is \( y = 12 + 18x - 5x^2 \), which is in the standard quadratic form. Here, \( a = -5 \), \( b = 18 \), and \( c = 12 \) are constants. The negative value of \( a \) tells us that the graph of this parabola opens downward, indicating that it will have a maximum point.
The solutions from the quadratic function are key to many engineering, physics, and business problems, where you often need to find maximum or minimum values.

The derivative of a function gives us the rate of change of the function with respect to one of its variables, often represented as \( y' \) or \( \frac{dy}{dx} \). For a quadratic function like \( y = 12 + 18x - 5x^2 \), the derivative provides a linear representation of our original function’s behavior.

Finding the Derivative

To find the derivative of \( y \), apply the power rule to each term:

• The derivative of 12 is 0, as constants have no slope.
• The derivative of \( 18x \) is 18, since the exponent 1 becomes the coefficient.
• The derivative of \( -5x^2 \) is \( -10x \), following the rule that brings the power in front and subtracts one from the exponent.

Thus, the derivative \( y' = 18 - 10x \). This equation helps us find the slope of the original function at any point \( x \).
To solve optimization problems, set the derivative equal to zero to find critical points where the function may reach a maximum or minimum.

The vertex of a parabola is the point where the curve changes direction, marking the highest or lowest point of the parabola, depending on its orientation. For a quadratic function \( y = ax^2 + bx + c \) that opens upwards or downwards, the vertex formula is crucial for finding this point.
For our given function \( y = 12 + 18x - 5x^2 \), it opens downward, indicating the vertex will provide a maximum point. The formula to find the \( x \)-coordinate of the vertex is given by \( x = -\frac{b}{2a} \).
Plugging in the values from our function, we have:\[x = -\frac{18}{2(-5)} = 1.8\]This \( x \)-value tells us where on the x-axis the vertex lies. To find the \( y \)-value of the vertex, substitute \( x = 1.8 \) back into the original equation:\[y = 12 + 18(1.8) - 5(1.8)^2 = 28.2\]This coordinates \( (1.8, 28.2) \) represent the vertex, confirming it is the maximum point in the given parabola.

The second derivative test helps confirm whether a critical point found using the first derivative is a maximum, minimum, or a point of inflection. For a function \( y \), the second derivative is represented as \( y'' \) or \( \frac{d^2y}{dx^2} \).
To apply the second derivative test, calculate \( y'' \) first. For our function \( y = 12 + 18x - 5x^2 \), we previously found \( y' = 18 - 10x \). Differentiating this again gives us:\[y'' = -10\]The constant \(-10\) is the second derivative, indicating a constant downward curvature without any change as \( x \) changes.

• If \( y'' < 0 \), the function is concave down, confirming that the critical point is a maximum.
• If \( y'' > 0 \), the function would have been concave up, indicating a minimum.

In our exercise, \( y'' = -10 \) implies the function is always concave down, ensuring that the point at \( x = 1.8 \) is indeed a maximum point. Using this test provides additional verification of our findings from the first derivative.