Problem 64 - College Algebra

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 equation is \( f(x) = 5x^2 - 77 \).

Step 1: Identify the General Form

A quadratic function with the vertex on the y-axis is commonly expressed in the form \( f(x) = ax^2 + c \). Given that the quadratic must have the same shape as \( f(x) = 5x^2 \), we know that \( a = 5 \). So the function will be \( f(x) = 5x^2 + c \).

Step 2: Use the Given Point to Find c

The point \( (4, 3) \) lies on the function, which means that when \( x = 4 \), \( f(x) = 3 \). Substitute these values into the equation \( f(x) = 5x^2 + c \) to find \( c \): \[ 3 = 5(4)^2 + c \] \[ 3 = 80 + c \] \[ c = 3 - 80 = -77 \].

Step 3: Write the Final Equation

With \( a = 5 \) and \( c = -77 \), substitute back into the quadratic function \( f(x) = ax^2 + c \). The equation of the function is \( f(x) = 5x^2 - 77 \).

Key Concepts

Vertex Form of a Quadratic

When working with quadratic functions, one popular representation is the vertex form. This form is expressed as \( f(x) = a(x - h)^2 + k \) where

• \( a \) determines the "width" and direction of the parabola (whether it opens upwards or downwards).
• \( (h, k) \) is the vertex of the parabola.

Understanding this form is helpful in visualizing how the function will look graphically. In the problem, however, the vertex is on the \( y \)-axis, which simplifies our function to exclude \( h \), assuming the vertex is at \( (0, k) \). This leads to the simpler form \( f(x) = ax^2 + c \) where \( c \) functions as \( k \) in our simplified vertex form.

Finding Coefficients

To solve quadratic equations, you might need to find unknown coefficients to fully define the function. These coefficients primarily impact the shape and position of the parabola. Given that the quadratic function has the shape of \( f(x) = 5x^2 \) we already know that \( a = 5 \). This tells us the parabola opens upwards and is relatively narrow due to the value of \( a \).
To find the coefficient \( c \), we use a given point on the function. For the function to pass through the given point \( (4, 3) \), we substitute this into the equation \( f(x) = 5x^2 + c \) so:

• \( 3 = 5(4)^2 + c \)
• \( 3 = 80 + c \)
• Solving this gives \( c = -77 \)

Thus, we have our function: \( f(x) = 5x^2 - 77 \).

Given Points in a Function

Using specific points that a quadratic function passes through is crucial when determining unknown coefficients. These points provide exact values to solve for those coefficients in an equation. In this exercise, the point \( (4, 3) \) gives us concrete numbers to substitute into \( f(x) = 5x^2 + c \) to find \( c \).
Points on a function are essentially solutions to the equation when \( x \) is substituted in, and the function equals \( y \). For example, substituting \( x = 4 \) and \( y = 3 \) helps us solve:

• Set \( 3 = 5(4)^2 + c \); note how the point is our \( (x, y) \).
• Simplifying gives \( 3 = 80 + c \).
• Finally, solve for \( c \) by evaluating \( c = 3 - 80 \).

By using these given points, you can tailor your function to meet specific conditions, offering a precise and meaningful quadratic equation.