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Chapter 1: Problem 1

In Problems 1-22, sketch the graph of each function. Do not use a graphing calculator. (Assume the largest possible domain.) $$ y=x^{2}+1 $$

Short Answer

Expert verified
The graph is an upward-opening parabola with vertex at \((0, 1)\) and axis of symmetry \(x = 0\).

Step by step solution

01

Identify the Function Type

The given function is a quadratic function, which generally has the form \(y = ax^2 + bx + c\). In this case, \(a = 1\), \(b = 0\), and \(c = 1\). The graph of a quadratic function is a parabola.
02

Determine the Vertex

For a quadratic function \(y = ax^2 + bx + c\), the vertex form is \(y = a(x-h)^2 + k\). Since \(b = 0\) in this function, the vertex is at \((0, c)\). So, the vertex of \(y = x^2 + 1\) is \((0, 1)\). This point is the lowest point on the graph because the parabola opens upwards (as \(a > 0\)).
03

Determine the Axis of Symmetry

The axis of symmetry for the parabola given by \(y = ax^2 + bx + c\) is the vertical line \(x = -\frac{b}{2a}\). Here, \(b = 0\) and \(a = 1\), so the axis of symmetry is \(x = 0\). This line divides the parabola into two mirror-image halves.
04

Identify Additional Points

To sketch the graph, it's helpful to find additional points. Compute \(y\) values for a few \(x\) values:- If \(x = 0\), \(y = (0)^2 + 1 = 1\).- If \(x = 1\), \(y = (1)^2 + 1 = 2\).- If \(x = -1\), \(y = (-1)^2 + 1 = 2\).- If \(x = 2\), \(y = (2)^2 + 1 = 5\).- If \(x = -2\), \(y = (-2)^2 + 1 = 5\).Plot these points: \((0, 1), (1, 2), (-1, 2), (2, 5), (-2, 5)\).
05

Sketch the Graph

Plot the vertex \((0, 1)\) and the other points calculated in Step 4. Draw a smooth, symmetrical curve, a parabola, through these points. The graph is upward opening with the vertex at its lowest point due to the positive \(a\) value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Parabolas
In mathematics, a parabola is a symmetrical open plane curve formed by the graph of a quadratic function. A quadratic function can generally be expressed in the form \( y = ax^2 + bx + c \). The graph of such a function is known as a parabola. Parabolas can open upwards or downwards depending on the sign of the coefficient \( a \):
  • If \( a > 0 \), the parabola opens upwards, resembling a U shape.
  • If \( a < 0 \), it opens downwards, similar to an upside-down U.
For the function \( y = x^2 + 1 \), \( a = 1 \) indicates that the parabola opens upwards. Parabolas have distinct properties like symmetry and a vertex which we will explore in the upcoming sections.
Identifying the Vertex
The vertex is an important feature of a parabola. It represents the highest or lowest point on the graph, depending on the orientation of the parabola. For the standard quadratic function \( y = ax^2 + bx + c \), the vertex can be determined using the formula: \( \left( -\frac{b}{2a}, f \left( -\frac{b}{2a} \right) \right) \). In the specific case of \( y = x^2 + 1 \):
  • Here, \( a = 1 \) and \( b = 0 \), which simplifies to the vertex being at \((0, c)\).
  • Since \( c = 1 \), the vertex is at the point \( (0, 1) \).
As \( a = 1 \) is positive, the vertex is the lowest point of this upward-opening parabola.
Explaining the Axis of Symmetry
A parabola is symmetric, and its axis of symmetry is a vertical line that passes through the vertex, dividing the parabola into two mirror-image halves. The equation for the axis of symmetry of a parabola given by \( y = ax^2 + bx + c \) is \( x = -\frac{b}{2a} \). Let's apply this to \( y = x^2 + 1 \):
  • With \( b = 0 \) and \( a = 1 \), the axis of symmetry is at \( x = 0 \).
This line \( x = 0 \) is a crucial guide when sketching the parabola as it ensures that any point \( (x, y) \) on one side of the axis has a corresponding point \( (-x, y) \) on the other side.
Key Points in Graph Sketching
Graph sketching involves a few straightforward steps to visualize the quadratic function effectively. After identifying the vertex and the axis of symmetry, here鈥檚 how you sketch \( y = x^2 + 1 \): Begin by plotting the vertex at \( (0, 1) \). From there, calculate additional points by substituting small integer values of \( x \) into the equation:
  • For \( x = 1 \) and \( x = -1 \), we find \( y = 2 \), giving points \( (1, 2) \) and \( (-1, 2) \).
  • For \( x = 2 \) and \( x = -2 \), we find \( y = 5 \), resulting in points \( (2, 5) \) and \( (-2, 5) \).
Plot these points and draw a smooth curve through them that opens upwards, ensuring symmetry about the line \( x = 0 \). With these steps, you will have accurately sketched a parabola.

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