Problem 66 - Algebra and Trigonometry

Question

Without graphing, determine whether the quadratic function \(f(x)=-3 x^{2}+12 x+5\) has a maximum value or a minimum value, and then find the value.

Step-by-Step Solution

Answer

The function has a maximum value of 17.

Step 1 - Determine the direction of the parabola

Identify the coefficient of the quadratic term, which is \( -3 \). Since it is negative, the parabola opens downward, indicating that the function has a maximum value.

Step 2 - Find the vertex

The vertex of a parabola given by the quadratic equation \(f(x)=ax^{2}+bx+c\) can be found using the formula for the x-coordinate of the vertex, \(x = -\frac{b}{2a}\). Here, \(a = -3 \) and \(b = 12 \), so \(x = -\frac{12}{2(-3)} = 2\).

Step 3 - Compute the value of the function at the vertex

Substitute \(x = 2\) back into the original function to find the y-coordinate: \(f(2) = -3(2)^{2}+12(2)+5\). Simplify it step by step: \(-3(4) + 24 + 5 = -12 + 24 + 5 = 17\).

Step 4 - Conclude the maximum value

Since the parabola opens downward, the y-coordinate of the vertex is the maximum value. Therefore, the maximum value of the function is \(17\).

Key Concepts

Parabola

A parabola is a U-shaped curve that can either open upwards or downwards. It is the graph of a quadratic function, which takes the form:

\[ f(x) = ax^2 + bx + c \]
Where:

• \(a\), \(b\), and \(c\) are constants,
• \(x\) represents the variable.

The direction in which the parabola opens depends on the sign of the coefficient \(a\):

• If \(a > 0\), the parabola opens upwards.
• If \(a < 0\), the parabola opens downwards.

In the given quadratic function, \(-3x^2 + 12x + 5,\) the coefficient of \(x^2\) is \(-3\). Because it is negative, the parabola opens downward. This tells us that the function has a maximum value, not a minimum.

Vertex Formula

The vertex of a parabola is its highest or lowest point, depending on the direction it opens. For the quadratic function \(f(x) = ax^2 + bx + c\), the vertex can be found using the vertex formula:
\[ x = -\frac{b}{2a} \]
Here,

• \(a\) is the coefficient of \(x^2\),
• \(b\) is the coefficient of \(x\).

In our function, \(a = -3\) and \(b = 12\). Plug these values into the vertex formula to get:
\[ x = -\frac{12}{2(-3)} = -\frac{12}{-6} = 2 \]
This tells us the x-coordinate of the vertex is 2. To find the y-coordinate, substitute \(x = 2\) back into the original function:
\[ f(2) = -3(2)^2 + 12(2) + 5 \]
Simplify step-by-step:
\[ f(2) = -3(4) + 24 + 5 \]
\[ f(2) = -12 + 24 + 5 \]
\[ f(2) = 17 \]
So, the vertex of the parabola is at \((2, 17)\), and 17 is the function's maximum value.

Maximum Value

In a quadratic function, the maximum or minimum value occurs at the vertex. For a parabola that opens downward (which occurs when \(a < 0\)), the vertex represents the maximum value.

In the function, \(f(x) = -3x^2 + 12x + 5\), since the parabola opens downward, it has a maximum value. We already found the vertex to be \((2, 17)\).

Therefore, the highest point on the parabola, and thus the maximum value of the function, is 17. This maximum value is the y-coordinate of the vertex. So, no matter what values you plug into the function, it will never output a number greater than 17.