91Ó°ÊÓ

Differentiation is a critical tool in calculus that allows us to find the rate at which a function is changing at any given point. When we differentiate a function, we derive its slope or gradient function, which tells us the slope of the tangent line to the curve at a specific point. This is immensely useful in understanding how functions behave and change. In the context of this exercise, we began by differentiating the quadratic function given:

• The original function: \( y = 4x^2 + 12x + 17 \).
• Using the power rule, \( rac{dy}{dx} = 8x + 12 \), gives the derivatives of each term: \( 8x \) from \( 4x^2 \), and \( 12 \) from \( 12x \). Constants disappear since their rate of change is zero.

Differentiation helps identify how steeply the function rises or falls at any specific value of \( x \). This leads to the discovery of critical points, important for determining maximum or minimum points on the graph of a quadratic function.

A quadratic function is a type of polynomial function that is defined as a function of the form \( ax^2 + bx + c \), where \( a, b, \) and \( c \) are constants and \( a eq 0 \). The graph of a quadratic function is a parabola, which can open upwards or downwards. Here are key characteristics to remember:

• The coefficient \( a \) determines the direction of the parabola's opening. If \( a > 0 \), it opens upwards; if \( a < 0 \), it opens downwards.
• The vertex of the parabola is the highest or lowest point, providing valuable information about the graph's shape and location.
• The vertex form of a quadratic function is \( y = a(x-h)^2 + k \), where \( (h, k) \) is the vertex of the parabola.

In this particular exercise, we find that the quadratic function is \( y = 4x^2 + 12x + 17 \), which opens upwards because 4 is positive. The vertex, found through differentiation, provides a minimum point for this parabola.

Critical points are where the first derivative of a function equals zero or is undefined. These points are essential in calculus because they help us find where a function reaches its maximum or minimum values, known as extrema. In the steps provided in this exercise, to locate this critical point:

• We set the derivative to zero: \( 8x + 12 = 0 \). This identifies the \( x \)-value where the slope of the tangent is zero, indicating a change in direction of the function.
• Solving \( 8x + 12 = 0 \) gives \( x = -\frac{3}{2} \). This \( x \)-value is a potential minimum or maximum point of the function.
• To determine its nature, we substitute back into the original function to find the corresponding \( y \)-value, which is part of the vertex.

In this scenario, substituting back provides the point \((-\frac{3}{2}, 8)\), signifying the parabolic function's vertex and its minimum point.