Problem 4 - Algebra and Trigonometry

Question

\(1-4\) . The graph of a quadratic function \(f\) is given. (a) Find the coordinates of the vertex. (b) Find the maximum or minimum value of \(f\) $$ f(x)=3 x^{2}+6 x-1 $$

Step-by-Step Solution

Answer

The vertex is at \((-1, -4)\), and the minimum value is \(-4\).

Step 1: Identify the Quadratic Function Components

We start by recognizing that the quadratic function is in the standard form, which is \( f(x) = ax^2 + bx + c \). Here, \( a = 3 \), \( b = 6 \), and \( c = -1 \). Since \( a > 0 \), the parabola opens upwards, indicating a minimum vertex.

Step 2: Find the Vertex using the Vertex Formula

To find the vertex of the parabola, we use the formula for the vertex \( x \)-coordinate \( x = -\frac{b}{2a} \). Substituting the values, we get \( x = -\frac{6}{2 \times 3} = -1 \).

Step 3: Calculate the Vertex \( y \)-coordinate

Substitute \( x = -1 \) into the function to find the \( y \)-coordinate: \( f(-1) = 3(-1)^2 + 6(-1) - 1 = 3 - 6 - 1 = -4 \). Therefore, the vertex is \((-1, -4)\).

Step 4: Determine the Minimum Value of \( f \)

Since the parabola opens upwards (\( a > 0 \)), the vertex represents the minimum point of the function. Thus, the minimum value of \( f(x) \) is the \( y \)-coordinate of the vertex, which is \(-4\).

Key Concepts

vertex of a parabola

In a quadratic function, the vertex is a crucial point where the curve changes direction. You can think of it as the peak or the lowest point of the parabola. This vertex can be found using the vertex formula, which relies on the coefficients from the standard form of the function. Remember, the standard form is represented as \( f(x) = ax^2 + bx + c \).

• The x-coordinate of the vertex is determined by the formula \( x = -\frac{b}{2a} \).
• The y-coordinate is found by substituting the x-coordinate back into the function.

As demonstrated in the original solution, once you calculate \( x = -1 \), you substitute \( x = -1 \) into \( f(x) = 3(-1)^2 + 6(-1) - 1 \) to get \( y = -4 \). Therefore, the vertex is at \((-1, -4)\). This point is important because it provides significant information about the function's graph.

minimum value

When discussing quadratic functions, the concept of minimum and maximum values is vital. For parabolas, whether they have a minimum or a maximum depends on the coefficient \( a \).

• If \( a > 0 \), the parabola opens upwards and has a minimum value.
• If \( a < 0 \), the parabola opens downwards and has a maximum value.

In our example, since \( a = 3 \) which is positive, the parabola opens upwards, meaning it has a minimum value. This minimum value is simply the y-coordinate of the vertex, which we've calculated as \(-4\).
Understanding where the minimum or maximum occurs helps in understanding the nature and behavior of the function. This minimum or maximum point tells us the lowest or highest outcome of the function.

standard form of a quadratic equation

The standard form of a quadratic equation is fundamental for understanding parabolas and their properties. It takes the form \( f(x) = ax^2 + bx + c \) where \( a \), \( b \), and \( c \) are constants. Each part of the standard form plays a specific role:

• \(a\): Determines the direction and width of the parabola.
• \(b\): Affects the position of the vertex along the x-axis.
• \(c\): Represents the y-intercept, where the parabola crosses the y-axis.

In our problem, with the function \( f(x) = 3x^2 + 6x - 1 \), we have \( a = 3 \), \( b = 6 \), and \( c = -1 \).
Knowing the standard form helps us not only find the vertex but also understand how slight changes in \( a \), \( b \), or \( c \) can drastically change the graph's shape and position. This understanding is key in both graphing and analyzing quadratic functions efficiently.