Problem 43 - Precalculus Mathematics for Calculus

Question

Find a function whose graph is a parabola with vertex \((1,-2)\) and that passes through the point \((4,16)\)

Step-by-Step Solution

Answer

The function is \( y = 2(x-1)^2 - 2 \).

Step 1: Understand the Parabola's Vertex Form

A parabola with a vertex can be expressed in vertex form: \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola. Here, the vertex is \((1, -2)\), so the equation starts as \( y = a(x-1)^2 - 2 \).

Step 2: Substitute the Known Point into the Equation

We know the parabola passes through the point \((4, 16)\). Substitute \(x = 4\) and \(y = 16\) into the vertex form equation: \( 16 = a(4-1)^2 - 2 \).

Step 3: Simplify the Equation

Simplify the expression: \( 16 = a(3)^2 - 2 \) which simplifies to \( 16 = 9a - 2 \).

Step 4: Solve for \(a\)

Add 2 to both sides: \( 18 = 9a \). Then, divide both sides by 9 to find \(a\): \( a = 2 \).

Step 5: Write the Function Equation

Substitute \(a = 2\) back into the vertex form equation: \( y = 2(x-1)^2 - 2 \). This is the equation of the parabola with the given vertex and point.

Key Concepts

Vertex Form

The vertex form of a parabola is an incredibly useful way to write quadratic equations, especially when you know the vertex of the parabola. The formula is: \( y = a(x-h)^2 + k \), where

• \((h, k)\) is the vertex of the parabola.
• The value \(a\) influences the direction and width of the parabola.

The vertex form is particularly beneficial because it immediately tells you the vertex point, which makes it easy to graph the parabola and understand its shape.
For instance, if the vertex is at \((1, -2)\), the formula becomes \( y = a(x-1)^2 - 2 \). This reflects a parabola that opens upwards if \(a > 0\) and downwards if \(a < 0\). The vertex form not only helps you identify the vertex quickly but also allows you to derive other characteristics of the parabola with ease.

Solving for a Constant

When you have a known point that the parabola passes through in addition to the vertex, you can extend the information to find out the constant \(a\). This process is called "solving for a constant."
Imagine you are given a point \((x, y)\) on the parabola, say \((4, 16)\). You can plug this point into the vertex form of the equation: \[ 16 = a(4 - 1)^2 - 2 \].
By performing algebraic manipulations, you simplify the equation: \[ 16 = 9a - 2 \]. The next step involves isolating \(a\). Add 2 to both sides to obtain \[ 18 = 9a \] and then divide by 9 to find \(a = 2\). This constant \(a\) describes how "steep" or "flat" the parabola is based on how it affects the width of the curve.

Substituting Points into Equations

In algebra, substituting points into equations is a way to use known variables to solve for unknowns. This technique works well when you have both the vertex of a parabola and another point on the curve.

• Begin by taking the coordinates from the point, here it’s \((4, 16)\), and substituting \(x = 4\) and \(y = 16\) into your vertex equation: \[ y = a(x-1)^2 - 2 \].
• Replace the appropriate values: \[ 16 = a(4-1)^2 - 2 \].

This substitution is a straightforward method to find unknown measurements, such as the value of \(a\). It involves simple arithmetic steps that transform the more complex equation into a manageable format for solving.