Problem 28 - College Algebra

Question

For the following exercises, use the vertex \((h, k)\) and a point on the graph \((x, y)\) to find the general form of the equation of the quadratic function. $$ (h, k)=(0,1),\left(x_{2} y\right)=(2,5) $$

Step-by-Step Solution

Answer

The quadratic function is \( f(x) = x^2 + 1 \).

Step 1: Understand the Vertex Form

The vertex form of a quadratic equation is given by \( f(x) = a(x-h)^2 + k \). This equation provides a way to express a quadratic function in terms of its vertex \( (h, k) \).

Step 2: Substitute Known Values into Vertex Form

The vertex provided is \((h, k)=(0,1)\). Substitute these values into the vertex form equation: \( f(x) = a(x-0)^2 + 1 \) which simplifies to \( f(x) = ax^2 + 1 \).

Step 3: Use the Given Point to Solve for 'a'

We are also given a point \((x, y) = (2, 5)\). Substitute \(x = 2\) and \(y = 5\) into the equation \( f(x) = ax^2 + 1 \) resulting in \( 5 = a(2)^2 + 1 \).

Step 4: Solve the Equation for 'a'

Simplify the equation \( 5 = 4a + 1 \) by subtracting 1 from both sides to get \( 4 = 4a \). Finally, divide by 4 to find \( a = 1 \).

Step 5: Write the General Form of the Quadratic Equation

With \( a = 1 \), substitute back into the vertex form: \( f(x) = 1x^2 + 1 \). Thus, the general form of the quadratic function is \( f(x) = x^2 + 1 \).

Key Concepts

Vertex Form

The vertex form of a quadratic equation is an important way to express a quadratic function. It is typically written as \( f(x) = a(x-h)^2 + k \). In this equation:

• \( a \) is a constant that affects the width and direction of the parabola.
• \( (h, k) \) is the vertex of the parabola, representing the point at which it is the lowest or highest.

In vertex form, the main advantage is that the vertex of the parabola is immediately visible, making it easy to graph and understand the parabola's position on the coordinate plane. For example, in the exercise, the vertex \(h,k\) is given as \( (0,1) \), which tells us the parabola's turning point is at this position.

General Form

Quadratic functions are often expressed in general form, which is written as \( ax^2 + bx + c \). This form is helpful for solving mathematical problems using algebraic techniques, such as factoring or using the quadratic formula. Each term in the equation provides significant information:

• \( a \) determines the direction of the parabola (upward if positive, downward if negative) and its width.
• \( b \) and \( c \) impact the position and the roots of the parabola on the graph.

In the step-by-step solution, the general form was derived from the vertex form. By substituting back the known value of "a" into the vertex equation, and completing simple algebraic manipulations, the equation \( f(x) = x^2 + 1 \) was found, which is a simple representation in general form.

Quadratic Equation

Quadratic equations are fundamental expressions in algebra that represent a polynomial of degree two. They typically appear in the form \( ax^2 + bx + c = 0 \), where the solutions—also known as roots—can be found using various methods:

• Factoring (if possible).
• Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Completing the square.

These equations are crucial since they appear in many real-world applications, such as physics problems, engineering challenges, and more. In the exercise, the quadratic function \( f(x) = x^2 + 1 \) describes a parabola that opens upward, but without a linear term \( bx \), making it simpler to work with.

Vertex

The vertex of a quadratic function is a key feature that represents its peak or trough. It lies at the axis of symmetry of the parabola. In vertex form \( f(x) = a(x-h)^2 + k \), the vertex is given by \((h, k)\):

• \( h \) is the x-coordinate.
• \( k \) is the y-coordinate.

The vertex can be a maximum or minimum point on the graph, depending on the direction of the parabola. If \( a > 0 \), the parabola opens upwards, and the vertex is a minimum point. Conversely, if \( a < 0 \), the parabola opens downwards, and the vertex is a maximum point. The given exercise presented the vertex \((0,1)\), making it easy to identify the center of the parabola as this is the turning point of the function.