Problem 28 - A Graphical Approach to College Algebra

Question

For each quadratic function defined , (a) use the vertex formula to find the coordinates of the vertex and (b) graph the function. Do not use a calculator. $$y=-x^{2}+2 x+1$$

Step-by-Step Solution

Answer

The vertex is at (1, 2) and the parabola opens downwards.

Step 1: Identify Coefficients

In the quadratic equation given by \( y = ax^2 + bx + c \), identify the coefficients. Here, \( a = -1 \), \( b = 2 \), and \( c = 1 \).

Step 2: Calculate Vertex using Vertex Formula

The vertex formula for a parabola given by \( y = ax^2 + bx + c \) is \( x_v = -\frac{b}{2a} \). Substitute the values of \( a \) and \( b \) to find \( x_v \): \[ x_v = -\frac{2}{2(-1)} = 1 \].

Step 3: Find the y-coordinate of the Vertex

Substitute \( x_v = 1 \) back into the original equation to find \( y_v \). Calculate \( y = -(1)^2 + 2(1) + 1 \): \( y = -1 + 2 + 1 = 2 \). Thus, the vertex of the function is \( (1, 2) \).

Step 4: Graph the Function

Graph the quadratic function \( y = -x^2 + 2x + 1 \) by plotting the vertex \((1, 2)\) and determining the direction of the parabola. Since \( a = -1 \), the parabola opens downwards. Include additional points such as the y-intercept \((0, 1)\) and symmetry point \((2, 1)\) to help draw the graph.

Key Concepts

Vertex Formula

The vertex of a quadratic function is a key point that helps us understand the graph's shape and direction. For quadratic equations of the form \( y = ax^2 + bx + c \), the vertex can be found using the vertex formula. This formula is given as \( x_v = -\frac{b}{2a} \).

This gives us the x-coordinate of the vertex. Once you have the x-coordinate, you can plug it back into the quadratic equation to calculate the y-coordinate, forming the complete vertex \((x_v, y_v)\). Let's take the quadratic function \( y = -x^2 + 2x + 1 \) as an example.

Using the vertex formula, we substitute \( a = -1 \) and \( b = 2 \) to obtain \( x_v = -\frac{2}{2(-1)} = 1 \). Next, we substitute \( x_v = 1 \) back into our equation to find \( y_v \). Calculating \( y = -(1)^2 + 2(1) + 1 = 2 \), we have the vertex as \((1,2)\). This point becomes crucial when sketching the graph or analyzing the function.

Graphing Parabolas

Once you have identified the vertex of the quadratic function, you can begin graphing the parabola. The vertex \((x_v, y_v)\) serves as the parabola's highest or lowest point, depending on its opening direction. In our example, we have the vertex \((1, 2)\).

Graphing typically starts with plotting this vertex on the coordinate plane. Then, identify additional points by selecting x-values near the vertex and calculating their corresponding y-values. These points will reveal the parabolic curve. In our quadratic function \( y = -x^2 + 2x + 1 \), points such as the y-intercept \((0, 1)\) and the symmetry point \((2, 1)\) are useful.

A parabola is symmetric relative to a vertical line passing through its vertex, known as the axis of symmetry. For our example, this line is \( x = 1 \). Using this symmetry helps in accurately drawing the parabola, ensuring both sides mirror each other.

Parabolas Opening Direction

The direction in which a parabola opens depends on the coefficient \( a \) in the quadratic function \( y = ax^2 + bx + c \). Understanding this coefficient's role is crucial in determining the graph's orientation
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If \( a > 0 \), the parabola opens upwards, making the vertex its lowest point. Conversely, if \( a < 0 \), the parabola opens downwards, making the vertex the highest point. In our given function \( y = -x^2 + 2x + 1 \), \( a = -1 \), which indicates that the parabola opens downwards.

This understanding of direction helps us predict the behavior of the graph and identify solutions to problems such as finding maximum or minimum values. Additionally, recognizing which way the parabola opens assists in determining how the variable values change as they approach the vertex. In essence, the sign of \( a \) plays a pivotal role in understanding the fundamental nature of quadratic functions.