Problem 64 - College Algebra with Corequisite Support

Question

For the following exercises, write the equation of the quadratic function that contains the given point and has the same shape as the given function. Contains (4,3) and has the shape of \(f(x)=5 x^{2}\). Vertex is on the \(y\) - axis.

Step-by-Step Solution

Answer

The quadratic function is \( g(x) = 5x^2 - 77 \).

Step 1: Understanding the given shape

Given the function shape as \( f(x) = 5x^2 \), we understand that the quadratic function we need to find has a similar form, \( g(x) = 5(x - h)^2 + k \), but we are given it has the vertex on the \( y \)-axis, implying \( h = 0 \). Thus, the equation simplifies to \( g(x) = 5x^2 + k \).

Step 2: Substitute the given point

The point given is \((4, 3)\). Substitute this point into the equation \( g(x) = 5x^2 + k \), resulting in \( 3 = 5(4)^2 + k \).

Step 3: Solve for k

From the previous step, we have \( 3 = 5(16) + k \), which simplifies to \( 3 = 80 + k \). By solving for \( k \), we find \( k = 3 - 80 = -77 \).

Step 4: Write the final function

Substitute \( k = -77 \) back into the equation \( g(x) = 5x^2 + k \), giving us the quadratic function \( g(x) = 5x^2 - 77 \).

Key Concepts

Vertex Form

When it comes to quadratic functions, one important form to know is the vertex form. The vertex form of a quadratic equation is generally written as \( g(x) = a(x-h)^2 + k \). Here, \( a \) determines the width and direction of the parabola, \( h \) is the x-coordinate of the vertex, and \( k \) is the y-coordinate of the vertex. In our specific problem, we verified that the vertex is located on the y-axis. This information tells us that \( h = 0 \), which means our equation reduces to \( g(x) = a(x-0)^2 + k \) or simply \( g(x) = ax^2 + k \).
Understanding this structure helps to identify parts of the parabola equation quickly and figure out the position and shape of the graph. Keeping the vertex form in mind can make it much easier to solve quadratic equations, especially when translating from a function's standard form to its graph or vice versa.

Function Transformation

Transforming quadratic functions involves a series of manipulations that shift, stretch or mirror the graph. In our example, the function \( f(x) = 5x^2 \) is the base function. To transform this base into another function with a particular point or vertex, there are key things to keep in mind:

• A horizontal shift happens when you change \( h \), which shifts the function left or right depending on its sign.
• A vertical shift is determined by \( k \), moving the function up or down.
• The \( a \) value will stretch or compress the parabola and determine if it opens upwards or downwards.

Since the parallel function to \( f(x) = 5x^2 \) has its vertex on the y-axis in this problem, there is no horizontal shift involved (\( h = 0 \)).We simply have a vertical transformation which is dictated by the \( k \) value. By substituting the given point into the equation, we find that \( k = -77 \). Thus, the final transformation results in the function \( g(x) = 5x^2 - 77 \), indicating a downward shift of the parabola by 77 units.

Equation Solving

Solving quadratic equations is critical in many mathematical contexts. In the given problem, we needed to find the exact equation that passes through a particular point and has a specific form. The point given was \( (4, 3) \) and the task was to incorporate this data into the quadratic formula.
Here’s how this was performed:

• The base function was known: \( f(x)=5x^2 \). We needed it in a similar form: \( g(x)=5x^2+k \).
• We substituted the known point: replace \( x \) with 4 and \( g(x) \) with 3, turning the equation into \( 3=5(4)^2+k \).
• By solving for \( k \), the expression developed: \( k = 3 - 80 \).
• The solution for \( k \) was \( -77 \), leading to the refined function: \( g(x) = 5x^2 - 77 \).

This process illustrates the use of known points to learn unknown elements of the equation, thereby crafting an accurate quadratic function.