Problem 31 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$f(x)=-x^{2}+2 x+5$$

Step-by-Step Solution

Answer

The vertex of the function is at (1, 6), the y-intercept is at (0, 5), and the x-intercepts can be determined by solving the equation \( -x^{2} + 2x + 5 = 0 \). The function is a downward-opening parabola, which can be sketched by plotting the vertex, x-intercepts, and y-intercept, and drawing a u-shaped curve that passes through these points.

Step 1: Identify the coefficients

From \( f(x) = -x^{2} + 2x + 5 \), we identify \( a = -1 \), \( b = 2 \), and \( c = 5 \).

Step 2: Calculate the Vertex

The coordinates of the vertex are given by \( h = -\frac{b}{2a} \) and \( k = f(h) \). Thus, \( h = -\frac{2}{2\times -1} = 1 \) and \( k = f(1) = -(1)^2 + 2(1) + 5 = 6 \). So, the vertex is \( (1,6) \).

Step 3: Find the y-intercept

The y-intercept is the y-coordinate when \( x = 0 \). Substituting \( x = 0 \) into the equation gives the y-intercept as \( f(0) = -(0)^2 + 2(0) + 5 = 5 \). So, \( y = 5 \) is the y-intercept.

Step 4: Find the x-intercepts

The x-intercepts can be found by setting \( f(x) = 0 \), which results in the quadratic equation \( -x^2 + 2x + 5 = 0 \). Solving this equation gives the x-coordinate of the x-intercepts.

Step 5: Sketch the Graph

Plot the vertex at (1, 6), the y-intercept at (0, 5), and the x-intercepts obtained from step 4. Connect these points to create a u-shaped curve that opens downward (since \( a < 0 \)). This sketches the graph of \( y = f(x) \).

Key Concepts

Vertex

The vertex of a quadratic function is a key feature of its graph. For the function given, which is in the form of a downward-opening parabola, the vertex represents the highest point on the graph. The vertex can be calculated using the formula: \( h = -\frac{b}{2a} \) for the x-coordinate and \( k = f(h) \) for the y-coordinate. In our function, \( f(x) = -x^2 + 2x + 5 \), the coefficients are \( a = -1 \), \( b = 2 \), and \( c = 5 \). By plugging in these values, we find that \( h = -\frac{2}{2 \times -1} = 1 \). When we substitute \( x = 1 \) back into the function, we get \( f(1) = -(1)^2 + 2(1) + 5 = 6 \). Thus, the vertex is at the point \((1, 6)\). This vertex helps us understand that the graph reaches its maximum at the point \( x = 1 \). It is an essential feature when graph sketching.

Intercepts

Intercepts are points where the graph intersects the axes. These include the y-intercept and the x-intercepts.

• The y-intercept is found by evaluating the function at \( x = 0 \). Plugging in the values, we get \( f(0) = -0^2 + 2 \times 0 + 5 = 5 \). Thus, the y-intercept is the point \((0, 5)\).
• X-intercepts occur where \( f(x) = 0 \). Solving the equation \( -x^2 + 2x + 5 = 0 \) reveals these points. This step typically involves using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For our equation, this calculation provides the x-intercepts, which are points where the parabola crosses the x-axis.

Understanding these intercepts is crucial as they give critical information about the graph's path across the axes.

Graph Sketching

Graph sketching involves plotting the curve by considering features such as the vertex and intercepts. In the case of \( f(x) = -x^2 + 2x + 5 \), the critical points include:

• The vertex at \((1, 6)\).
• The y-intercept at \((0, 5)\).
• The x-intercepts determined from solving the quadratic equation.

These points help to define the parabolic shape. The parabola is open downward because the coefficient \( a = -1 \) is negative, indicating that it arcs toward the downward direction. A well-sketched graph of this function will smoothly connect all these critical points, reflecting a symmetrical curve centered around the vertex. Remember, symmetrical placement around the vertex is a distinguishing feature of parabolas.

Quadratic Equation

A quadratic equation is a second-degree polynomial of the form \( ax^2 + bx + c = 0 \). The given function \( f(x) = -x^2 + 2x + 5 \) is a common example of a quadratic function, where \( a = -1 \), \( b = 2 \), and \( c = 5 \). Quadratic equations can be solved in various ways:

• Factoring: When possible, factoring allows for simpler solutions. However, not all quadratics are easily factored.
• Quadratic Formula: This universal method uses the formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which is applicable for any quadratic.
• Completing the Square: This technique involves rewriting the equation in a form that makes it easier to spotlight its vertex.Each of these approaches gives unique insights into the behavior of the quadratic. Understanding the structure and solution methods of quadratic equations is crucial, as they frequently appear in both academic settings and real-world applications.