Problem 95 - Precalculus : Building Concepts and Connections

Question

A parabola associated with a certain quadratic function \(f\) has the point (2,8) as its vertex and passes through the point \((4,0) .\) Find an expression for \(f(x)\) in the form \(f(x)=a(x-h)^{2}+k\) (a) From the given information, find the values of \(h\) and \(k\) (b) Substitute the values you found for \(h\) and \(k\) into the expression \(f(x)=a(x-h)^{2}+k\) (c) Now find \(a\). To do this, use the fact that the parabola passes through the point \((4,0) .\) That is, \(f(4)=0\) You should get an equation having just \(a\) as a variable. Solve for \(a\) (d) Substitute the value you found for \(a\) into the expression you found in part (b). (e) Graph the function using a graphing utility and check your answer. Is (2,8) the vertex of the parabola? Does the parabola pass through (4,0)\(?\)

Step-by-Step Solution

Answer

The expression for the quadratic function \(f\) in vertex form is \(f(x)=-2 (x-2)^{2}+8\).

Step 1: Determine \(h\) and \(k\)

\The coordinates of the vertex of the parabola are given as (2,8). In vertex form, these correspond to the values of \(h\) and \(k\), respectively. Therefore, \(h=2\) and \(k=8\).

Step 2: Substitute \(h\) and \(k\)

Substitute \(h\) and \(k\) into the vertex form of the parabola \(f(x)=a (x-h)^{2}+k\). This gives: \(f(x)=a (x-2)^{2}+8\).

Step 3: Determine \(a\)

Use the point \((4,0)\) on the parabola to determine \(a\). Substitute \(x=4\) and \(f(4)=0\) into the formula leading to \(0=a (4-2)^{2}+8 \). Simplifying the right side of the equation gives \(0=4a + 8\). Solving for \(a\) gives \(a = -2\).

Step 4: Substitute \(a\)

Having found \(a\), substitute it back into the formula from step 2. This gives the function \(f(x)=-2 (x-2)^{2}+8\).

Step 5: Graph and check

Finally, using a graphing utility graph the function \(f(x)=-2 (x-2)^{2}+8\). Check whether \((2,8)\) is the vertex and if the parabola passes through \((4,0) \).

Key Concepts

Vertex Form

The vertex form of a quadratic function is a very useful representation for understanding the geometric properties of a parabola. Given a quadratic function in the form \[f(x) = a(x-h)^2 + k\]- The vertex \((h, k)\) represents the highest or lowest point of the parabola, depending on the orientation.- The value \(a\) determines whether the parabola opens upwards \((a > 0)\) or downwards \((a < 0)\).This form is particularly beneficial when graphing because you directly know where the vertex of the parabola is located.
For example, if a parabola has its vertex at (2,8), then in vertex form, we set \(h = 2\) and \(k = 8\). Therefore, the beginning structure of your vertex form will look like this: \[f(x) = a(x-2)^2 + 8\]
Solving for \(a\) will help complete the function representation in vertex form.

Parabola

A parabola is a curved, symmetrical plane graph that is the graph of a quadratic function. It exhibits a unique property: every parabola is defined by its apex, known as the vertex.- Parabolas can either "open" upwards or downwards. This orientation is guided by the sign of the coefficient \(a\) in the quadratic function's vertex form. \(a > 0\) results in an upwards opening while \(a < 0\) causes a downwards opening.- In the given problem, since \(a=-2\), the parabola opens downward.
Furthermore, every parabola has an axis of symmetry which is a vertical line passing through its vertex. For the vertex (2,8), this line is \(x=2\).
As the parabola graphed using \(f(x)=-2(x-2)^2+8\) suggests, the vertex acts as a peak point because of the downward opening. That's why the function's maximum value is 8, which occurs when \(x=2\).

Graphing Utilities

Graphing utilities are powerful tools that allow one to visualize the shape and position of functions like parabolas efficiently. With the help of graphing utilities, students can gain a deeper understanding of function behavior and key features.

• Graphing calculators or software like Desmos can be used to plot quadratic functions like \(f(x) = -2(x-2)^2 + 8\).
• These tools provide visual confirmation that the function's vertex coincides with the point (2,8) and that the parabola touches or crosses other expected points, such as (4,0) in our example.
• They can also help explore transformations such as translations, dilations, and reflections.

When graphing, you start by entering the function's formula into the graphing utility. This will instantly display its parabola, giving insights into significant points, slopes, and the graph's orientation. Such visualization ensures clarity, especially in confirming the solution steps.