Problem 2 - Calculus for Biology and Medicine

Question

sketch the graph of each function. Do not use a graphing calculator. (Assume the largest possible domain.) $$ y=-(x-2)^{2}+1 $$

Step-by-Step Solution

Answer

A downward-opening parabola with vertex at (2, 1), axis of symmetry x = 2, and intercepts (0, -3), (1, 0), and (3, 0).

Step 1: Identify the Parent Function

The given function is in the form of the standard quadratic function \( y = ax^2 + bx + c \). The parent function is \( y = x^2 \), which is a parabola with its vertex at the origin \((0,0)\), opening upwards.

Step 2: Determine the Transformations

The function \( y = -(x-2)^2 + 1 \) indicates a series of transformations to the parent function. The inside of the parenthesis, \( x-2 \), tells us there is a horizontal shift. This represents a shift to the right by 2 units. The negative sign in front of \((x-2)^2\) causes a reflection over the x-axis, making the parabola open downward. Finally, the \(+1\) indicates a vertical shift upward by 1 unit.

Step 3: Find the Vertex

From the transformation \( x-2 \), the vertex of the parabola is shifted right to \( x = 2 \). With the \(+1\) vertical shift, the vertex is located at \((2, 1)\). The function is reflected over the x-axis, so the vertex will be the highest point.

Step 4: Determine the Axis of Symmetry

The axis of symmetry of a parabola in the form \( y = -(x-h)^2 + k \) is a vertical line passing through the vertex. For our function \( y = -(x-2)^2 + 1 \), this line is \( x = 2 \).

Step 5: Sketch Intersections

To find where the parabola intersects the y-axis, set \( x = 0 \): \( y = -(0-2)^2 + 1 = -4 + 1 = -3 \). The intersection point is \((0, -3)\). The x-intercepts occur where \( y = 0 \): \( 0 = -(x-2)^2 + 1 \). Solving gives \( (x-2)^2 = 1 \), leading to \( x-2 = 1 \) or \( x-2 = -1 \), which gives intersection points \((3, 0)\) and \((1, 0)\).

Step 6: Sketch the Graph

Plot the vertex \((2, 1)\), the y-intercept \((0, -3)\), and the x-intercepts \((3, 0)\) and \((1, 0)\). Draw the parabola opening downward from the vertex, passing through these points, maintaining symmetry about the axis \( x = 2 \).

Key Concepts

Vertex Form of a Quadratic Equation

Understanding quadratic functions becomes simpler when we express them in what is called the "vertex form". This form is given by \( y = a(x-h)^2 + k \), where \( (h, k) \) is the vertex of the parabola.
This vertex is crucial because it tells us the 'peak' or 'valley' of the parabola. The variable \( a \) affects the direction and the "width" of the parabola. When \( a \) is positive, the parabola opens upwards, and when negative, it opens downwards.
For the function \( y = -(x-2)^2 + 1 \), it is already in vertex form. Here, the vertex point is \((2, 1)\), which means the highest point since the parabola opens downward due to the negative sign before \( (x-2)^2 \).
The vertex form is a friendly format because it readily gives you the vertex, making it easier to visualize how the parabola is placed on the graph.

Reflections and Transformations

Quadratic functions undergo various transformations that alter their shape or position on a graph.
With our function \( y = -(x-2)^2 + 1 \), there are a few transformations to note:

• A horizontal shift caused by \( x-2 \), moving the graph to the right by 2 units.
• A reflection over the x-axis, indicated by the negative sign in front of \((x-2)^2\). This flips the parabola so it opens downwards instead of upwards.
• A vertical shift upwards by 1 unit, represented by the \(+1\). This moves the whole graph up along the y-axis.

Each transformation plays an important role. Horizontal shifts change the starting point along the x-axis, reflections adjust the overall direction, and vertical shifts raise or lower the graph. Understanding these can help you plot the most accurate graph.

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. It goes straight through the vertex.
In vertex form, the axis of symmetry is easily spotted since it is \( x = h \).
For our equation \( y = -(x-2)^2 + 1 \), the vertex is \((2, 1)\), making the axis of symmetry \( x = 2 \).
This means the parabola is a mirror image on either side of this vertical line. The axis of symmetry helps to understand how the graph is symmetrically balanced and aids in plotting accurate points on either side of it.

Intercepts of Quadratic Functions

Intercepts are the points where the graph crosses the axes.
For the y-intercept, set \( x = 0 \) in the equation \( y = -(x-2)^2 + 1 \). This gives us \( y = -3 \), meaning the y-intercept is the point \( (0, -3) \).
For x-intercepts, set \( y = 0 \). Solving \( 0 = -(x-2)^2 + 1 \) gives two x-intercepts. These are \( x = 3 \) and \( x = 1 \), so the points are \((3, 0)\) and \((1, 0)\).
These intercepts are crucial as they help in sketching a precise graph, showing where and how the parabola crosses the axes. Remember, knowing the intercepts allows a sketch that is more than just dots and lines—it's an insightful visualization.