Problem 40 - Beginning and Intermediate Algebra

Question

Rewrite function in the form \(f(x)=a(x-h)^{2}+k\) by completing the square. Then, graph the function. Include the intercepts. \(h(x)=-\frac{1}{2} x^{2}-3 x-\frac{19}{2}\)

Step-by-Step Solution

Answer

The given function can be rewritten as \(h(x) = -\frac{1}{2}(x+3)^2 - \frac{1}{2}\) by completing the square. The vertex is \((-3, -\frac{1}{2})\), and the axis of symmetry is \(x = -3\). To find the x-intercepts, solve \(0 = -\frac{1}{2}(x+3)^2 - \frac{1}{2}\) (you might need to use the quadratic formula). The y-intercept is \(h(0) = -\frac{13}{2}\). Use this information to graph the function.

Step 1: Complete the Square for the given function

First, we write the given function as: \(h(x) = -\frac{1}{2}x^2 - 3x - \frac{19}{2}\) Now, let's complete the square for the quadratic part of the expression \(-\frac{1}{2}x^2 - 3x\) \(h(x) = -\frac{1}{2}(x^2 + 6x) - \frac{19}{2}\) To complete the square, take the coefficient of the linear term in the parenthesis (6 in this case), divide it by 2 and square the result. In our case, it's \((\frac{6}{2})^2=9\). Now add and subtract the obtained value inside the parenthesis, and factor the the resulting perfect square trinomial. \(h(x) = -\frac{1}{2}(x^2 + 6x + 9 - 9) - \frac{19}{2}\) Now, rewrite the expression \(h(x) = -\frac{1}{2}((x^2 + 6x + 9) - 18) - \frac{19}{2}\) Now, we factor out the perfect square trinomial \(h(x) = -\frac{1}{2}((x + 3)^2 - 18) - \frac{19}{2}\) Now distribute the \(-\frac{1}{2}\) term. \(h(x) = -\frac{1}{2}(x+3)^2 + 9 - \frac{19}{2}\) Finally, combine the constants \(h(x) = -\frac{1}{2}(x+3)^2 + \frac{-1}{2}\)

Step 2: Find the x and y intercepts

To find the x-intercepts, we need to set \(h(x) = 0\) and solve for x: \(0 = -\frac{1}{2}(x+3)^2 + \frac{-1}{2}\) To find the y-intercept, we need to set \(x = 0\) and solve for h(x): \(h(0) = -\frac{1}{2}(0+3)^2 + \frac{-1}{2}\)

Step 3: Graph the function

Using the completed square form, we can identify the vertex, axis of symmetry, and intercepts to create a graph of the function. Vertex: \((-3, -\frac{1}{2})\) Axis of symmetry: \(x = -3\) X-intercepts: Solve \(0 = -\frac{1}{2}(x+3)^2 + \frac{-1}{2}\) (You might need to use the quadratic formula if manual factoring does not work) Y-intercept: \(h(0) = -\frac{1}{2}(3)^2 +\frac{-1}{2}=-\frac{13}{2}\) Now, use this information to graph the function.

Key Concepts

Quadratic Functions

Quadratic functions are essential in mathematics because they represent parabolic graphs. They are generally polynomial functions of degree 2, and the standard form is given by:

• The equation: \( f(x) = ax^2 + bx + c \)
• The letter 'a' is non-zero.
• The parameter 'b' is a linear coefficient.
• The constant 'c' is the y-intercept.

Quadratic functions have unique properties like symmetry and a maximum or minimum point called the vertex. This vertex is a significant feature used in graphing these functions. Moreover, quadratic functions can be transformed into different forms, such as vertex form, which offers easier insights into their graphical representation.

Graphing Functions

Graphing a quadratic function involves a few key steps. First, understanding the shape of the graph, known as a parabola, is crucial. Parabolas can open upwards or downwards, depending on the sign of the coefficient 'a'.

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

To graph a quadratic function from its vertex form, identifying the vertex and axis of symmetry is crucial.
Additionally, finding intercepts will help define the graph's position on the coordinate plane:

• The vertex gives the graph's maximum or minimum point.
• The axis of symmetry is a line passing through the vertex, splitting the parabola in half.
• X and y-intercepts provide specific points where the graph crosses the axes.

After identifying these points, plot them on the coordinate plane and sketch the parabola symmetrically around the straight axis of symmetry.

Vertex Form

The vertex form of a quadratic function is incredibly useful for graphing, offering a clear picture of the key features of a parabola:

• The general form is \( f(x) = a(x - h)^2 + k \)
• Here, \( (h, k) \) is the vertex of the parabola.
• The parameter 'a' determines the direction and width of the parabola.

Using the vertex form, graphing becomes simpler as the vertex point gives an immediate location on the graph. This reduced complexity is valuable when sketching graphs quickly. In the specific example from the exercise, the vertex is at \((-3, -\frac{1}{2})\), and the equation is in the form \(-\frac{1}{2}(x+3)^2 + \frac{-1}{2}\), providing direct insight into the parabola's graph.

Intercepts

The intercepts of a graph are the points where it crosses the x and y axes.

• The **x-intercepts** are found by setting the function equal to zero and solving for x. These points tell us where the graph will cross the horizontal axis.
• The **y-intercept** is found by setting \( x = 0 \) and evaluating the function. This tells us where the graph will intersect the vertical axis.

For the function \( h(x) = -\frac{1}{2}(x+3)^2 + \frac{-1}{2} \), the y-intercept can be calculated directly from the equation, resulting in \( -\frac{13}{2} \). Finding x-intercepts involves solving \( 0 = -\frac{1}{2}(x+3)^2 + \frac{-1}{2} \), possibly requiring methods like factoring or using the quadratic formula. These intercepts are essential in defining the spatial orientation of the graph on the coordinate plane, providing critical intersection points for understanding graph behavior.