Quadratic Functions: Videos & Practice Problems

Quadratic functions are a specific type of polynomial function characterized by their degree of 2. The standard form of a quadratic function is expressed as \( f(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are real numbers, and \( a \) cannot be zero. The graph of any quadratic function takes the shape of a parabola, which can open either upward or downward depending on the sign of \( a \).

The vertex of a parabola is a crucial feature, representing either the highest or lowest point on the graph. If the parabola opens upward, the vertex is a minimum point; if it opens downward, the vertex is a maximum point. The vertex is denoted as an ordered pair, such as \( (h, k) \), where \( h \) and \( k \) are the x and y coordinates, respectively.

Another important aspect of quadratic functions is the x-intercepts, which are the points where the graph crosses the x-axis. A quadratic function can have either one or two x-intercepts, but never more or less. The y-intercept, on the other hand, is where the graph intersects the y-axis, and it can be found by evaluating \( f(0) \).

The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves, always passing through the vertex. It can be expressed as \( x = h \), where \( h \) is the x-coordinate of the vertex.

When analyzing the domain and range of quadratic functions, the domain is always all real numbers, represented as \( (-\infty, \infty) \). The range, however, depends on whether the vertex is a minimum or maximum. For a minimum point, the range extends from the vertex's y-coordinate to infinity, while for a maximum point, it extends from negative infinity to the vertex's y-coordinate.

Additionally, understanding the intervals of increase and decrease is essential. A parabola increases on the interval from negative infinity to the vertex and decreases from the vertex to positive infinity. This behavior is determined by the vertex's position, which acts as a dividing point for these intervals.

Quadratic functions can also be expressed in vertex form, which is particularly useful for graphing. The vertex form is given by \( f(x) = a(x - h)^2 + k \), where \( (h, k) \) is the vertex of the parabola. This form allows for easier identification of the vertex and the direction in which the parabola opens.

Understanding the vertex form of a quadratic function is essential for graphing parabolas effectively. The vertex form is expressed as:

$$f(x) = a(x - h)^2 + k$$

In this equation, the parameters \(a\), \(h\), and \(k\) represent specific transformations of the basic quadratic function \(f(x) = x^2\). The value of \(a\) determines the direction and width of the parabola. If \(a\) is positive, the parabola opens upwards, while a negative \(a\) indicates it opens downwards. The absolute value of \(a\) also affects the shape: if \(|a| > 1\), the parabola undergoes a vertical stretch, making it narrower; if \(|a| < 1\), it experiences a vertical compression, resulting in a wider shape.

The parameter \(h\) indicates a horizontal shift. Since the equation is \(x - h\), a positive \(h\) shifts the graph to the right, while a negative \(h\) shifts it to the left. The parameter \(k\) represents a vertical shift, where a positive \(k\) moves the graph up and a negative \(k\) moves it down.

To graph a quadratic function in vertex form, follow these steps:

1. **Identify the Vertex**: The vertex of the parabola is given by the point \((h, k)\). This point is crucial as it represents the highest or lowest point of the graph, depending on the sign of \(a\). If \(a\) is positive, the vertex is a minimum; if negative, it is a maximum.

2. **Determine the Axis of Symmetry**: The axis of symmetry is a vertical line that passes through the vertex, expressed as \(x = h\). This line divides the parabola into two mirror-image halves.

3. **Find the X-Intercepts**: To find the x-intercepts, set \(f(x) = 0\) and solve for \(x\). This often involves rearranging the equation into a form suitable for applying the square root property or factoring.

4. **Calculate the Y-Intercept**: The y-intercept can be found by evaluating \(f(0)\). Substitute \(0\) for \(x\) in the vertex form equation and simplify to find the corresponding \(y\) value.

5. **Plot Points and Sketch the Graph**: With the vertex, axis of symmetry, x-intercepts, and y-intercept identified, plot these points on a coordinate plane. Connect them with a smooth curve to form the parabola, ensuring to extend the ends with arrows to indicate that the graph continues indefinitely.

By following these steps, you can accurately graph any quadratic function in vertex form, utilizing the transformations represented by \(a\), \(h\), and \(k\) to visualize the function's behavior effectively.

To graph the quadratic function f(x) = -\frac{1}{2}(x + 1)^2 + 2, we start by identifying key features of the parabola. The function is in vertex form, which is expressed as f(x) = a(x - h)^2 + k, where (h, k) is the vertex. In this case, we can rewrite the function to find the vertex:

Here, x + 1 can be interpreted as x - (-1), indicating that h = -1 and k = 2. Thus, the vertex is at the point (-1, 2). Since the coefficient a = -\frac{1}{2} is negative, the parabola opens downward, confirming that the vertex is a maximum point.

The axis of symmetry for the parabola is given by the line x = h, which in this case is x = -1.

Next, we find the x-intercepts by setting f(x) = 0:

- \frac{1}{2}(x + 1)^2 + 2 = 0

Rearranging gives:

- \frac{1}{2}(x + 1)^2 = -2

Multiplying both sides by -2 results in:

(x + 1)^2 = 4

Taking the square root of both sides yields:

x + 1 = \pm 2

Solving for x gives two x-intercepts: x = 1 and x = -3.

To find the y-intercept, we evaluate f(0):

f(0) = -\frac{1}{2}(0 + 1)^2 + 2 = -\frac{1}{2}(1) + 2 = -\frac{1}{2} + 2 = \frac{3}{2}

Thus, the y-intercept is (0, \frac{3}{2}).

Now we can summarize the key points for graphing:

• Vertex: (-1, 2) (maximum point)
• Axis of symmetry: x = -1
• X-intercepts: (1, 0) and (-3, 0)
• Y-intercept: (0, \frac{3}{2})

When graphing, plot the vertex, axis of symmetry, x-intercepts, and y-intercept. Connect these points with a smooth curve to form the parabola, which opens downward.

For the domain of the quadratic function, it is always (-\infty, \infty), indicating that all real numbers are included. The range, however, is limited by the maximum point at the vertex, so it is (-\infty, 2], where 2 is included in the range.

Lastly, the intervals of increase and decrease can be determined from the graph. The function is increasing on the interval (-\infty, -1) and decreasing on the interval (-1, \infty). The vertex at x = -1 is not included in these intervals as it represents the transition point.

When working with quadratic functions, it's essential to understand how to convert a function from standard form to vertex form, especially when you want to identify the vertex and intercepts for graphing. The vertex form of a quadratic function is expressed as \( f(x) = a(x - h)^2 + k \), where \( (h, k) \) represents the vertex of the parabola. If you start with a quadratic in standard form, such as \( f(x) = ax^2 + bx + c \), you can transform it into vertex form by completing the square.

To illustrate this process, consider the function \( f(x) = x^2 + 6x + 7 \). First, identify the coefficients: \( a = 1 \), \( b = 6 \), and \( c = 7 \). The first step is to factor out \( a \) from the first two terms, which in this case remains \( x^2 + 6x \) since \( a = 1 \).

Next, to complete the square, calculate \( \frac{b}{2a} \). Here, \( \frac{6}{2 \cdot 1} = 3 \). Squaring this value gives \( 3^2 = 9 \). You then add and subtract this square inside the parentheses: \( x^2 + 6x + 9 - 9 + 7 \). This manipulation allows you to maintain the equality of the function while preparing to factor.

Now, move the subtracted value outside the parentheses, accounting for the factor of \( a \). Since \( a = 1 \), this step simplifies to just subtracting 9 from the constant term. The expression now reads \( (x^2 + 6x + 9) - 2 \), which can be factored as \( (x + 3)^2 - 2 \).

At this point, the function is in vertex form: \( f(x) = (x + 3)^2 - 2 \). From this, you can easily identify the vertex \( (h, k) = (-3, -2) \). This transformation not only allows for easier graphing but also provides a clear understanding of the parabola's characteristics, such as its direction and vertex position.

In summary, converting from standard form to vertex form involves factoring, completing the square, and rearranging terms. This method is a powerful tool for analyzing quadratic functions and graphing them effectively.

To analyze the quadratic function f(x) = -2x² - 4x + 6, we will convert it from standard form to vertex form by completing the square, which simplifies the graphing process. The coefficients are identified as a = -2, b = -4, and c = 6.

First, we factor out a from the first two terms:

f(x) = -2(x² + 2x) + 6

Next, we calculate b/(2a) to complete the square. Here, b = -4 and a = -2, so:

b/(2a) = -4/(2 * -2) = 1

We then add and subtract 1 inside the parentheses:

f(x) = -2(x² + 2x + 1 - 1) + 6

Rearranging gives:

f(x) = -2((x + 1)² - 1) + 6

Distributing the -2 results in:

f(x) = -2(x + 1)² + 2 + 6

Thus, we have:

f(x) = -2(x + 1)² + 8

Now, we can identify the vertex of the parabola, which is given by the coordinates (h, k). From the vertex form, h = -1 and k = 8, indicating that the vertex is at (-1, 8). Since the coefficient of the squared term is negative, the parabola opens downwards, making the vertex a maximum point.

The axis of symmetry is the vertical line x = h = -1.

To find the x-intercepts, we set f(x) = 0:

-2(x + 1)² + 8 = 0

Rearranging gives:

-2(x + 1)² = -8

Dividing by -2 results in:

(x + 1)² = 4

Taking the square root of both sides yields:

x + 1 = ±2

Thus, we find:

x = -1 + 2 = 1 and x = -1 - 2 = -3

The x-intercepts are (1, 0) and (-3, 0).

Next, we calculate the y-intercept by evaluating f(0):

f(0) = -2(0 + 1)² + 8 = -2(1) + 8 = 6

Thus, the y-intercept is (0, 6).

Now we can summarize the key features of the parabola:

• Vertex: (-1, 8) (maximum point)
• Axis of Symmetry: x = -1
• X-Intercepts: (1, 0) and (-3, 0)
• Y-Intercept: (0, 6)
• Domain: (−∞, ∞)
• Range: (−∞, 8]
• Increasing Interval: (−∞, -1)
• Decreasing Interval: (-1, ∞)

With this information, we can graph the parabola, ensuring to plot the vertex, intercepts, and axis of symmetry, and connect these points with a smooth curve to represent the downward-opening parabola accurately.