Problem 28 - Algebra A Combined Function

Question

Find the vertex of the graph of each quadratic function. Determine whether the graph opens upward or downward, find any intercepts, and graph the function. $$ f(x)=-4 x^{2}+8 x $$

Step-by-Step Solution

Answer

The vertex is (1, 4), the graph opens downward, intercepts are (0, 0) and (2, 0).

Step 1: Identify the coefficients

The given quadratic function is \( f(x) = -4x^2 + 8x \). The general form of a quadratic function is \( ax^2 + bx + c \). Here, \( a = -4 \), \( b = 8 \), and \( c = 0 \).

Step 2: Determine the direction of opening

The sign of the coefficient \( a \) determines whether the parabola opens upwards or downwards. Since \( a = -4 \) is negative, the graph opens downward.

Step 3: Calculate the vertex

The vertex \((h, k)\) can be found using the formula \( h = -\frac{b}{2a} \). Substituting the values \( b = 8 \) and \( a = -4 \), we have \( h = -\frac{8}{2(-4)} = 1 \). To find \( k \), substitute \( x = 1 \) back into the function: \( k = f(1) = -4(1)^2 + 8(1) = 4 \). So, the vertex is \((1, 4)\).

Step 4: Find the intercepts

**x-intercepts**: Set \( f(x) = 0 \). Thus, \(-4x^2 + 8x = 0 \). Factoring gives \(-4x(x - 2) = 0 \). The solutions are \( x = 0 \) and \( x = 2 \). These are the x-intercepts. **y-intercept**: Set \( x = 0 \). Thus, \( f(0) = -4(0)^2 + 8(0) = 0 \). Hence, the y-intercept is \((0, 0)\).

Step 5: Graph the function

With the vertex \((1, 4)\) and intercepts at \((0, 0)\) and \((2, 0)\), plot these points on the coordinate plane. The parabola opens downward, so the curve extends down through these points, with \((1, 4)\) as the highest point.

Key Concepts

Vertex of a Parabola

In quadratic functions, the vertex represents an important point. It is either the highest or lowest point, depending on the curve's nature. For our specific function, which is quadratic and in the standard form, the vertex can be found using the formula: \[ h = -\frac{b}{2a} \].
By substituting the given values, \( b = 8 \) and \( a = -4 \), we find that \( h = 1 \).
To find the \( k \) value of the vertex, we substitute \( x = 1 \) back into the function. This results in \( k = 4 \).

• The vertex is located at \((1, 4)\).
• This is the point where the direction of the curve changes.

Direction of Parabolic Opening

The direction in which a parabola opens is determined by the coefficient \( a \) in the quadratic function.
If \( a \) is positive, the parabola opens upwards; if \( a \) is negative, it opens downward.
In this quadratic, \( a = -4 \), which is a negative value. This tells us that the parabola opens downwards. This means the vertex is the highest point on the graph.

• A negative \( a \) means the parabola makes a 'frowny face' shape.
• The downward opening indicates the vertex is a maximum point.

Intercepts of a Quadratic Function

Intercepts are points where the graph crosses the axes.

• x-intercepts: These occur where the function equals zero. For this function, setting \( -4x^2 + 8x = 0 \) and factoring gives us \( x = 0 \) or \( x = 2 \). So, the x-intercepts are \((0, 0)\) and \((2, 0)\).
• y-intercept: The y-intercept is found by setting \( x = 0 \), resulting in the point \((0, 0)\).

These intercepts are important as they help define the curve's placement in the coordinate system.

Graphing Quadratic Functions

Graphing a quadratic function involves plotting its vertex, intercepts, and the general shape dictated by the direction of opening.
Start by marking the vertex \((1, 4)\), which is the curve's peak for downward opening parabolas.
Next, plot the x-intercepts \((0, 0)\) and \((2, 0)\) along with the y-intercept which is the same as one of the x-intercepts here, \((0, 0)\).

• The shape will be symmetrical around the vertex line \( x = 1 \).
• The points give you the foundation to draw the parabola.
• Ensure the arms of the parabola extend downwards, reflecting the negative \( a \).

Graphing helps visualize the function and see how its components come together.