Problem 26 - Algebra and Trigonometry

Question

A quadratic function is given. (a) Express the quadratic function in standard form. (b) Sketch its graph. (c) Find its maximum or minimum value. $$ f(x)=5 x^{2}+30 x+4 $$

Step-by-Step Solution

Answer

The quadratic is in vertex form as \( f(x) = 5(x+3)^2 - 41 \), has a minimum value of \(-41\), and opens upwards with vertex at \((-3, -41)\).

Step 1: Identify the Quadratic's Parameters

The quadratic function is given by \( f(x) = ax^2 + bx + c \). In this function, \( a = 5 \), \( b = 30 \), and \( c = 4 \). Identifying these parameters will help in further steps, such as completing the square or using the vertex formula.

Step 2: Complete the Square to Find the Vertex

Convert the quadratic into vertex form, \( f(x) = a(x - h)^2 + k \). First, factor out the 5 from the first two terms: \[ f(x) = 5(x^2 + 6x) + 4. \] Now, complete the square inside the parentheses. Take half of 6 (which is 3), square it (which is 9), and add and subtract it inside the parenthesis:\[ f(x) = 5(x^2 + 6x + 9 - 9) + 4 = 5((x + 3)^2 - 9) + 4. \] So, \[ f(x) = 5(x + 3)^2 - 45 + 4 = 5(x + 3)^2 - 41. \] Thus, in vertex form, the function is \( f(x) = 5(x+3)^2 - 41 \), revealing the vertex \( (h, k) = (-3, -41) \).

Step 3: Sketch the Graph

The vertex form \( f(x) = 5(x+3)^2 - 41 \) indicates the vertex is at \((-3, -41)\) and opens upwards because \( a = 5 > 0 \). This means it has a minimum point at the vertex. The axis of symmetry is \( x = -3 \). The graph looks like a parabola opening upwards, with its lowest point at the vertex.

Step 4: Determine Maximum or Minimum Value

Since the quadratic has an upward-opening parabola (as \( a > 0 \)), it does not have a maximum value but instead has a minimum value at its vertex. The minimum value of \( f(x) \) is \(-41\). This value corresponds to the \( y \- ext{coordinate} \) of the vertex.

Key Concepts

Vertex Form

The vertex form of a quadratic function is a very useful way to express a quadratic equation as it easily reveals important properties of the graph. The vertex form is given by the equation:

• \[ f(x) = a(x - h)^2 + k \]

In this form,

• \( (h, k) \) represents the vertex of the parabola.
• The parabolic graph opens upwards if \( a > 0 \) and downwards if \( a < 0 \).

The vertex is the highest or lowest point on the graph. Converting a quadratic from standard form \( ax^2 + bx + c \) to vertex form is achieved through completing the square.Identifying the vertex gives insights into the graphical behavior of the quadratic. For instance, it tells you where the graph changes direction and what point is the peak or trough of the graph. To illustrate, for the expression we have converted, the vertex is at \( (-3, -41) \), which indicates the lowest point of the parabola.

Completing the Square

Completing the square is a process used to rewrite quadratic expressions so they are expressed in vertex form, making the vertex more easily identifiable.Here's a simple breakdown of the steps involved:

• Start with the quadratic in the form \( ax^2 + bx + c \).
• Factor out \( a \) from the x-terms (ignore the constant \( c \)). For instance, \( 5(x^2 + 6x) \).
• Take half of the coefficient of \( x \), square it, then add and subtract that value inside the bracket: \[f(x) = 5(x^2 + 6x + 9 - 9)\]
• Rearrange inside the parentheses to create a perfect square trinomial: \[f(x) = 5((x + 3)^2 - 9)\]
• Simplify further to achieve vertex form.

This process allows us to manipulate the quadratic into a form that highlights the vertex, helping to understand the nature and orientation of the graph.

Maximum and Minimum Values

In the realm of quadratic functions, understanding maximum and minimum values is crucial.For a quadratic expressed as \( ax^2 + bx + c \):

• If \( a > 0 \), the parabola opens upwards, suggesting the vertex is at the minimum point.
• If \( a < 0 \), it opens downwards, indicating the vertex as the maximum point.

The vertex determines these crucial points of extremum. In our converted function, the parabola opens upwards since \( a = 5 > 0 \), and thus it has a minimum value at its vertex.The minimum value of the function \( f(x) = 5(x + 3)^2 - 41 \) is the \( y \)-coordinate of the vertex, which is \(-41\). This value represents the smallest output that the function can achieve, and it is obtained at the point \( x = -3 \). Recognizing these values provides a deeper understanding of the behavior and graph of quadratic functions.