Problem 63 - 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 \((1,-3)\) and has the shape of \(f(x)=-x^{2}\) Vertex is on the \(y\) -axis.

Step-by-Step Solution

Answer

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

Step 1: Standard Form Identification

A quadratic function with its vertex on the y-axis and in the form similar to the function \( f(x) = -x^2 \) can be written as \( f(x) = a(x - h)^2 + k \), where \( h \) is the x-coordinate of the vertex. Since the vertex is on the y-axis, \( h = 0 \), and the equation simplifies to \( f(x) = a(x^2) + k \). We know that the function has the shape of \( f(x) = -x^2 \), so \( a = -1 \). Thus, the equation simplifies to \( f(x) = -(x^2) + k \).

Step 2: Substitute the Point into the Equation

To find \( k \), substitute the given point \((1, -3)\) into the equation. This gives \( -3 = -(1^2) + k \).

Step 3: Solve for Constant \(k\)

Calculate \( k \) by solving the equation from Step 2. Simplify \( -3 = -1 + k \), which gives \( k = -3 + 1 \) resulting in \( k = -2 \).

Step 4: Write the Final Equation

With \( a = -1 \) and \( k = -2 \), the final quadratic function is \( f(x) = -x^2 - 2 \). This function passes through the point \((1, -3)\) and has the same shape as \( f(x) = -x^2 \).

Key Concepts

Standard Form of a Quadratic

Quadratic functions are often presented in their standard form, which helps in visualizing and understanding their properties easily. The standard form of a quadratic equation is expressed as:

• \[ f(x) = a(x - h)^2 + k \]

Here,

• \(a\) indicates the direction and width of the parabola. A negative \(a\) opens downward, and a positive \(a\) opens upward.
• \(h\) and \(k\) are the coordinates of the vertex of the parabola, \((h, k)\).

Placing the quadratic in this standard form allows you to easily identify key features such as the vertex and the axis of symmetry. In problems where the vertex lies on the y-axis, \(h = 0\), simplifying our equation to \( f(x) = a(x^2) + k \). This makes recognizing the vertex a straightforward task.

Vertex on y-axis

When the vertex of a quadratic function is located on the y-axis, it simplifies our equation significantly. This means that the x-coordinate of the vertex, \(h\), is zero. Thus, our standard form becomes:

• \[ f(x) = a(x^2) + k \]

Because \(h = 0\), the axis of symmetry – the vertical line that divides the parabola into two equal halves – is the y-axis itself. This gives us a clearer view of its symmetrical properties. Additionally, knowing that the vertex lies on the y-axis can be particularly useful for graphing purposes, as it helps in defining the precise central curve of the parabola.

Point Substitution

Point substitution is a technique used to find unknown constants in a quadratic equation. By substituting a known point \((x, y)\) into the expression, we can solve for unknown parameters. Consider you have the equation:

• \[ f(x) = a(x^2) + k \]

Given a point \((1, -3)\), substitute

• \(x = 1\) and \(y = -3\)\ into the equation, getting \(-3 = a(1^2) + k\).

This process allows us to express \(k\) in terms of known and unknown values, paving the way for solving for constants. Point substitution is a crucial step in modeling and graph comprehension. It ensures the quadratic equation accurately represents all given conditions.

Solving for Constants

To solve for constants within a quadratic function, you focus on isolating the variable. Let's use the following equation obtained from point substitution:

• \[-3 = a(1^2) + k\]

Once you've substituted the point into the equation, it becomes easier to simplify and solve for \(k\). We know \(a = -1\) from the shape provided, which makes the equation

• \[-3 = -1 + k\]

To find \(k\), perform basic arithmetic: add 1 to both sides to isolate \(k\):

• \(k = -3 + 1 = -2\)

With our constants determined, you finalize the equation: \(f(x) = -x^2 - 2\). This formula now meets all initial criteria, passing through the point and matching the parabola's shape described in the problem.