Problem 22 - Algebra 2

Question

Graph each function. If \(a>0\) find the minimum value. If \(a<0\) find the maximum value. $$ y=-x^{2}+2 x+5 $$

Step-by-Step Solution

Answer

The maximum value of the function \(y=-x^{2}+2x+5\) is 6 at \(x=1\), and the graph is a downward-opening parabola with the vertex at \((1, 6)\).

Step 1: Identify the Type of Function

Notice that the given function, \(y=-x^{2}+2x+5\), is a quadratic function in the form \(y=ax^{2}+bx+c\), where \(a=-1\), \(b=2\), and \(c=5\).

Step 2: Find the Vertex

Since \(a=-1\), which is less than zero, the parabola opens downwards. To find the vertex of the parabola use the formula \(x = -\frac{b}{2a}\). For our function, \(x = -\frac{2}{2(-1)} = -\frac{2}{-2} = 1\). Then, compute the value of \(y\) by substituting \(x=1\) into the original equation: \(y = -1^2 + 2(1) + 5 = -1 + 2 + 5 = 6\). Hence, the vertex is \((1, 6)\).

Step 3: Find the Maximum Value

As \(a<0\), the quadratic function attains a maximum value at the vertex. Thus, the maximum value of the function is \(y=6\) at \(x=1\).

Step 4: Graph the Function

Plot the vertex and the y-intercept, which we can get by setting \(x=0\), resulting in \(y=c=5\). The parabola is symmetric with respect to the line \(x=1\). Draw the parabola opening downwards with the vertex at the highest point.

Key Concepts

Vertex of a Parabola

Understanding the vertex of a parabola is crucial when graphing quadratic functions. The vertex represents the highest or lowest point on the graph, depending on the direction the parabola opens. For the function \(y=-x^{2}+2x+5\), we can determine the vertex by using the formula \(x = -\frac{b}{2a}\).
For our example, \(a=-1\) and \(b=2\), so our vertex's x-coordinate is \(x = -\frac{2}{2(-1)} = 1\). Plugging this x-value into the original equation gives us the y-coordinate: \(y = -1^{2} + 2(1) + 5 = 6\). Hence, the vertex of this parabola is at the point \( (1, 6) \).

Understanding the Vertex Coordinates

To refine our understanding of the coordinates, it's important to remember that the x-coordinate of the vertex is the axis of symmetry for the parabola and the y-coordinate represents the maximum or minimum value the quadratic function can reach, which leads to our next topic.

Maximum Value of a Quadratic Function

The maximum value of a quadratic function is found at the vertex when the parabola opens downwards, which occurs if \(a<0\). This is the case with our function \(y=-x^{2}+2x+5\), where \(a=-1\). Since we've already calculated the vertex to be \( (1, 6) \), the maximum value is the y-coordinate of the vertex, which is 6.
This value is crucial for understanding the range of the function and predicting the maximum output for given inputs. It represents the peak value that the function can achieve.

Real-world Applications

In practical scenarios, like projectile motion in physics, the maximum value can represent the highest point an object reaches before descending. This illustrates the significance of being able to determine the maximum value in various contexts.

Parabola Symmetry

Symmetry in parabolas refers to the fact that they are mirror images on either side of a vertical line known as the axis of symmetry. For the function \(y=-x^{2}+2x+5\), the axis of symmetry is the line \(x=1\), which is directly derived from the vertex's x-coordinate. This property of symmetry is essential for graphing because it allows us to predict the shape and position of the parabola on the cartesian plane even with limited points.
By plotting the vertex and a few additional points, we can sketch the entire parabola through reflective symmetry across the axis. This quality also underpins many of the problem-solving strategies for quadratic equations, as it presents an element of predictability in an otherwise complex function.

Graphing Tips

When graphing by hand, after plotting the vertex, you can find additional points to the left or right of the axis and then mirror them across the axis for greater accuracy in your parabola's shape. Always remember that the parabola is a smooth curve and not a series of connected linear segments.