Problem 35 - Algebra and Trigonometry

Question

Find the maximum or minimum value of the function. $$ f(t)=100-49 t-7 t^{2} $$

Step-by-Step Solution

Answer

The maximum value of the function is -157.25.

Step 1: Identify the function type

The given function is a quadratic function of the form \( f(t) = at^2 + bt + c \), which is \( f(t) = -7t^2 - 49t + 100 \). This suggests that the function is a parabola.

Step 2: Determine the direction of the parabola

Since the coefficient of \( t^2 \) is \( -7 \), which is less than zero, the parabola opens downward. This means our quadratic function has a maximum value.

Step 3: Use the vertex formula to find the critical value

To find the t-coordinate of the vertex, use the formula \( t = -\frac{b}{2a} \). Here, \( a = -7 \) and \( b = -49 \). Substitute these values to find \( t = -\frac{-49}{2(-7)} \).

Step 4: Calculate the t-coordinate of the vertex

Compute \( t = \frac{49}{14} \), which simplifies to \( t = 3.5 \). This is the t-value at which the maximum point occurs.

Step 5: Find the maximum value of the function

Substitute \( t = 3.5 \) back into the function to get the maximum value: \( f(3.5) = 100 - 49(3.5) - 7(3.5)^2 \).

Step 6: Solve for the function's value at t = 3.5

Calculate: \( f(3.5) = 100 - 171.5 - 85.75 = -157.25 \). Hence, the maximum value of the function is \( -157.25 \).

Key Concepts

Vertex Formula

The vertex formula is a handy tool when working with quadratic functions like our problem where we have a function in the form of \( f(t) = at^2 + bt + c \). This formula allows us to find the **vertex** of the parabola represented by the function. The vertex is an essential point because it contains the maximum or minimum value of the function. To find the t-coordinate of the vertex, you use the formula: \[ t = -\frac{b}{2a} \] Here, \( a \) and \( b \) are coefficients from the quadratic equation. Plug these values into the formula to get the t-coordinate, which tells you where on the t-axis the maximum or minimum point happens. In our example, the vertex occurs at \( t = 3.5 \). Knowing this, you can then substitute this t-value back into the original function to find the actual maximum or minimum value. Understanding the vertex formula is crucial for solving problems involving quadratic functions efficiently.

Parabola

A parabola is a unique U-shaped curve described by quadratic functions like \( f(t) = -7t^2 - 49t + 100 \). Every quadratic function graphs into a parabola. The direction of the parabola—whether it opens up or down—is determined by the sign of the coefficient \( a \) in the quadratic term \( at^2 \).

• If \( a \) is positive, the parabola opens upward, resembling a bowl.
• If \( a \) is negative, as in our example, the parabola opens downward, like an umbrella.

In the context of our problem, since \( a = -7 \), the parabola opens downward. This means the vertex represents a maximum point because the parabola does not extend infinitely upwards. Understanding the shape and direction of a parabola helps us interpret and predict the behavior of quadratic functions.

Maximum Value

The maximum value of a quadratic function is the highest point on its graph when it forms a downward-opening parabola. In our exercise, once we found the t-coordinate of the vertex using the vertex formula, the next step was to substitute this t-value back into the original function. This calculation finds the function's value at the vertex. For \( t = 3.5 \), we plug into the function: \[ f(3.5) = 100 - 49(3.5) - 7(3.5)^2 \] After simplifying, we get \( f(3.5) = -157.25 \). This result is the maximum value of the function, as our parabola opens downward. Recognizing that this process determines the highest point on the graph is vital whenever dealing with downward-opening parabolas.

Quadratic Equation Solutions

Quadratic equations, which are instances of quadratic functions set equal to zero, can be solved to find their roots or solutions. These roots represent the points where the parabola intersects the t-axis, also known as the x-axis in generic terms. Although solving for roots was not necessary in our exercise about finding maximum or minimum values, it's crucial to understand their role in quadratic functions. The methods to solve these equations include:

• Factoring: Express the equation as a product of two binomials.
• Quadratic Formula: Use \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the solutions.
• Completing the Square: Rearrange the equation into a perfect square trinomial.

Each method has its strengths, and choosing the right one depends on the specific equation at hand. Grasping these solution techniques is essential for solving quadratic equations efficiently.