Problem 43 - Algebra and Trigonometry

Question

Find a function whose graph is a parabola with vertex and that passes through the point . $$ \begin{array}{l}{\text { Find a function whose graph is a parabola with vertex }(1,-2)} \\ {\text { and that passes through the point }(4,16) .}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Vertex Form of a Parabola

The vertex form of a parabola is given by the equation \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola. In this problem, the vertex is \((1, -2)\). Thus, the equation becomes \( y = a(x-1)^2 - 2 \).

Step 2: Substitute the Given Point into the Equation

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

Step 3: Simplify and Solve for the Parameter \(a\)

Simplify the equation from Step 2:\(16 = a(3)^2 - 2\)\(16 = 9a - 2\)Add 2 to both sides:\(18 = 9a\)Divide both sides by 9:\(a = 2\).

Step 4: Write the Final Function Equation

Substitute \(a = 2\) back into the vertex form equation:\( y = 2(x-1)^2 - 2 \).This is the equation of the parabola that has vertex \((1, -2)\) and passes through the point \((4, 16)\).

Key Concepts

Quadratic Function

A quadratic function is a type of polynomial function that is characterized by the highest power of the variable being squared, typically written in the form \( ax^2 + bx + c \). It is a foundational concept in algebra and is used to describe a parabolic curve when graphed. Understanding quadratic functions is crucial, as they appear in various real-world applications, from physics to engineering.

The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \), but there is also a vertex form that is particularly useful for identifying the parabola's vertex. The vertex form is given by \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex of the parabola.

• \(a\) determines the direction and "width" of the parabola
• \((h, k)\) gives the coordinates of the vertex
• When \(a > 0\), the parabola opens upwards
• When \(a < 0\), the parabola opens downwards

Understanding the vertex form can make it easier to identify the key features of the graph.

Parabola Graphing

Graphing a parabola involves plotting a quadratic function on a coordinate plane. The shape of the graph is a "U"-shaped curve called a parabola, which can open upwards or downwards.

There are several crucial points and aspects to note when graphing a parabola:

• Vertex: The parabola's highest or lowest point, given by \((h, k)\) in the vertex form.
• Axis of Symmetry: A vertical line that passes through the vertex, given by the equation \(x = h\).
• Direction of Opening: Determined by the sign of \(a\). If \(a > 0\), it opens upwards; if \(a < 0\), it opens downwards.
• Intercepts: Points where the parabola crosses the axes.

For example, using the function \( y = 2(x-1)^2 - 2 \), we can easily graph the parabola. It has a vertex at \((1, -2)\), and since \(a = 2\), it opens upward.

Solving Equations

Solving equations involving quadratic functions often means finding values of \(x\) that satisfy the quadratic equation. Here, we'll focus on solving for the parameter \(a\) in the vertex form of a parabola.

In our original exercise, we are tasked to find the function of a parabola with a given vertex and a point it passes through. We start by substituting the point into the vertex form,
then solve for \(a\).

• Start from the given vertex form equation: \( y = a(x - h)^2 + k \).
• Plug the given point \((x, y)\) into the equation.
• Solve for \(a\) using basic algebraic steps like simplifying, adding, and dividing.

In our case, substituting \((4, 16)\) into \(y = a(x-1)^2 - 2\) and solving gave us \(a = 2\).

This solved quadratic equation forms the basis of determining the specific parabola that fits the conditions of the problem.