Problem 35 - Precalculus : Building Concepts and Connections

Question

Write each quadratic function in the form \(f(x)=a(x-h)^{2}+k\) by completing the square. Also find the vertex of the associated parabola and determine whether it is a maximum or minimum point. $$w(x)=-x^{2}+6 x+4$$

Step-by-Step Solution

Answer

The given function written in vertex form is \(w(x) = -(x-3)^{2}+5\). The vertex of the parabola is (3, 5), and it represents a maximum point because the parabola opens downwards.

Step 1: Rewrite the Quadratic Function by Completing the Square

To complete the square for the function \(w(x)=-x^{2}+6x+4\), the constant term (4) needs to be temporarily removed. Divide the coefficient of the linear term by 2, square it, and add and subtract that quantity to obtain a perfect square trinomial in the quadratic equation. The function becomes \(w(x)=-[x^{2}-6x+9 -9]+4=-[(x-3)^{2}-9]+4\).

Step 2: Simplify to Vertex Form

Bring the added number back to the equation (with its sign) to simplify the equation to vertex form. Thus, \(w(x)=-[(x-3)^{2}-9]+4=-[(x-3)^{2}-5]=-(x-3)^{2}+5\). Therefore, the function in vertex form is \(w(x)=-(x-3)^{2}+5\).

Step 3: Identify the Vertex (h, k)

The quadratic function in vertex form \(f(x)=a(x-h)^{2}+k\) indicates the vertex of the parabola as (h, k). This gives us the vertex (3, 5).

Step 4: Determine whether the Vertex is a Maximum or Minimum point

The coefficient a in the function w(x) determines the direction in which the parabola opens. A positive a opens upwards, and a negative a opens downwards. Since our function has a coefficient of -1, the parabola opens downwards, which means that the vertex (3, 5) is a maximum point.

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic function into vertex form, which is particularly useful in identifying the vertex of its graph. To effectively complete the square for a given quadratic like \(w(x) = -x^2 + 6x + 4\), follow these simplified steps:

• First, factor out any coefficient of \(x^2\) if necessary. In our function, this involves temporarily removing the constant, leading to \(w(x) = -[x^2 - 6x] + 4\).
• Next, take half the coefficient of the \(x\) term (in this case, 6), which is 3, and square it, giving 9.
• Add and subtract this squared value inside the brackets: \(w(x) = -[x^2 - 6x + 9 - 9] + 4\).
• Express the trinomial as a perfect square: \(-(x-3)^2 - 9\).

Finally, rearrange and simplify to find the vertex form: \(w(x) = -(x-3)^2 + 5\). This represents our quadratic in vertex form.

Maximum and Minimum Points

Quadratic functions have a single highest or lowest point, known as a maximum or minimum point, determined by the vertex. In the function \(w(x) = -(x-3)^2 + 5\), it is crucial to identify if the vertex represents a maximum or a minimum.
The coefficient in front of the squared term, \(a\), guides this determination:

• If \(a > 0\), the parabola opens upwards, giving a minimum vertex.
• If \(a < 0\), the parabola opens downwards, signaling a maximum vertex.

In this case, the coefficient \(-1\) indicates a downward opening, confirming the vertex \((3, 5)\) as a maximum point. Thus, the highest point the parabola reaches on the graph is at \(y = 5\), which gives the maximum value of the function.

Parabola Vertex

The vertex of a parabola is the point on the graph that represents its highest or lowest point. In a quadratic function presented in vertex form, \(f(x) = a(x-h)^2 + k\), the vertex \((h, k)\) easily pinpoints this crucial point.

• From our function \(w(x) = -(x-3)^2 + 5\), the vertex is instantly identified as \((3, 5)\).
• This vertex location directly influences the symmetry of the parabola, being its reflection point.
• Knowing the vertex makes it effortless to sketch or analyze the parabola's path.

Vertex form not only simplifies finding the vertex but also helps track how transformations like shifts and stretches affect its positioning on the graph. This makes understanding and visualizing quadratic functions far more intuitive.