Problem 32 - 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. $$ h(x)=3-4 x-4 x^{2} $$

Step-by-Step Solution

Answer

The function in standard form is \(-4x^2 - 4x + 3\). The maximum value is 0 at \(x = \frac{1}{2}\).

Step 1: Identify the Given Quadratic Function

We start by identifying the given quadratic function: \( h(x) = 3 - 4x - 4x^2 \). This is not in the standard form yet. The standard form of a quadratic function is \( ax^2 + bx + c \).

Step 2: Rearrange into Standard Form

Reorder the terms of the quadratic function to match the standard form \( ax^2 + bx + c \). The given function \( h(x) = 3 - 4x - 4x^2 \) can be rewritten as \( h(x) = -4x^2 - 4x + 3 \). In this form, we have \( a = -4 \), \( b = -4 \), and \( c = 3 \).

Step 3: Sketch the Graph

To sketch the quadratic function, note that the coefficient \( a = -4 \) implies the parabola opens downward. We can find the vertex using the formula \( x = -\frac{b}{2a} \). Substitute \( a = -4 \) and \( b = -4 \):\[x = -\frac{-4}{2(-4)} = \frac{1}{2}\]Calculate \( h(x) \) at \( x = \frac{1}{2} \) to find the vertex's y-coordinate:\[h(\frac{1}{2}) = -4(\frac{1}{2})^2 - 4(\frac{1}{2}) + 3 = -1 - 2 + 3 = 0\]Thus, the vertex is \( (\frac{1}{2}, 0) \) and it's a maximum point. Sketch the parabola opening downward with this vertex and find the y-intercept by setting \( x = 0 \), getting \( h(0) = 3 \).

Step 4: Find the Maximum Value

Since the parabola opens downward, the vertex is the maximum point. The maximum value is the y-coordinate of the vertex, \( h(\frac{1}{2}) = 0 \).

Key Concepts

Standard Form of a Quadratic Function

Quadratic functions often start in a form that needs tweaking to reveal their secrets. The standard form of a quadratic function is given by \( ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants where \( a eq 0 \). This structure is neat because:

• It tells you whether the parabola opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)).
• It's straightforward for spotting coefficients directly affecting the curve's shape.

To transform a quadratic into standard form, rearrange all terms to match \( ax^2 + bx + c \). In our exercise, the function \( h(x) = 3 - 4x - 4x^2 \) is reshaped into \( h(x) = -4x^2 - 4x + 3 \).
Now, you have \( a = -4 \), \( b = -4 \), and \( c = 3 \). Step complete, and it's downhill from here! This format helps unveil more about the parabola's shape and position in a blink.

Vertex of a Parabola

The vertex of a parabola is like its 'tip' or 'peak.' For a parabola modeled by \( ax^2 + bx + c \), the vertex can act as either its lowest or highest point. The position of the vertex on the horizontal axis is found with a simple formula: \( x = -\frac{b}{2a} \).In this problem, substituting \( a = -4 \) and \( b = -4 \) into the formula gives: \[x = -\frac{-4}{2(-4)} = \frac{1}{2}.\]To find the vertex's y-coordinate, plug \( x = \frac{1}{2} \) back into the function:\[ h\left(\frac{1}{2}\right) = -4\left(\frac{1}{2}\right)^2 - 4\left(\frac{1}{2}\right) + 3 = 0.\]Thus, the vertex is \( (\frac{1}{2}, 0) \).

• It's the highest point (since \( a < 0 \)).
• Vital for sketching the curve and understanding its direction.

Understanding the vertex helps decode the path and peak of any parabola.

Maximum and Minimum Values of Quadratic Functions

For quadratic functions, the vertex holds the key to uncovering the max or min value. Whether it's a peak or a pit depends entirely on \( a \):

• If \( a > 0 \), the parabola opens upward with a minimum point at the vertex.
• If \( a < 0 \), it opens downward and the maximum point lies at the vertex.

In our exercise, we have \( a = -4 \), indicating a downward-open parabola. Hence, the vertex \( (\frac{1}{2}, 0) \) reveals the maximum value. \[\text{Maximum value} = 0\]Finding the maximum or minimum value can quickly indicate where a quadratic function's graph tops out or bottoms down, helping ground wider calculations and analysis.