Problem 294 - Calculus Volume 1

Question

For the following exercises, draw a graph of the functions without using a calculator. Be sure to notice all important features of the graph: local maxima and minima, inflection points, and asymptotic behavior. $$ y=3 x^{2}+2 x+4 $$

Step-by-Step Solution

Answer

The graph is an upward-opening parabola with vertex \((-\frac{1}{3}, \frac{25}{3})\) and y-intercept \((0, 4)\).

Step 1: Identify the Type of Function

The given function is a quadratic function of the form \( y = ax^2 + bx + c \). With \( a = 3 \), \( b = 2 \), and \( c = 4 \), this is a parabola opening upwards because the coefficient of \( x^2 \) is positive.

Step 2: Determine the Vertex

For a quadratic function \( y = ax^2 + bx + c \), the vertex \((h, k)\) can be found using the formula \( h = -\frac{b}{2a} \). Substituting \( b = 2 \) and \( a = 3 \), we get \( h = -\frac{2}{2 \times 3} = -\frac{1}{3} \). Substituting \( x = -\frac{1}{3} \) back into the function gives \( k = 3(-\frac{1}{3})^2 + 2(-\frac{1}{3}) + 4 = \frac{25}{3} \). So, the vertex is \((-\frac{1}{3}, \frac{25}{3})\).

Step 3: Find the Axis of Symmetry

The axis of symmetry for a parabola \( ax^2 + bx + c \) is a vertical line \( x = h \). From Step 2, \( x = -\frac{1}{3} \) is the axis of symmetry.

Step 4: Determine the Y-Intercept

The y-intercept of a function is the point where the graph intersects the y-axis, which occurs at \( x = 0 \). Substituting \( x = 0 \) into the equation, \( y = 3(0)^2 + 2(0) + 4 = 4 \). Therefore, the y-intercept is \((0, 4)\).

Step 5: Evaluate End Behavior

As \( x \to \pm \infty \), the term \( 3x^2 \) will dominate the behavior of the function, causing \( y \to \infty \). Therefore, the graph will rise to the top left and top right as it extends along the x-axis.

Step 6: Identifying Local Maxima or Minima

Since the parabola opens upwards and there are no other terms to influence multiple turnings, the vertex itself represents the parabola’s minimum point. Thus, the local minimum is \((-\frac{1}{3}, \frac{25}{3})\) and the function has no local maxima.

Step 7: Identify Inflection Points and Asymptotic Behavior

Quadratic functions like this one, given by \( y = ax^2 + bx + c \), do not have inflection points, as their curvature does not change sign. Also, there are no asymptotes as the parabola extends indefinitely in vertical directions.

Step 8: Sketch the Graph

Using all the information gathered: the vertex \((-\frac{1}{3}, \frac{25}{3})\), the y-intercept \((0, 4)\), and the symmetry axis \( x = -\frac{1}{3} \), sketch a parabola opening upwards with these features. Ensure the parabola is symmetric about \( x = -\frac{1}{3} \) and passes through \((0, 4)\).

Key Concepts

Parabola Vertex

The vertex of a parabola is a key concept in understanding quadratic functions. A vertex is essentially the highest or lowest point on the graph of a parabola. This depends on whether the parabola opens upwards or downwards. For a quadratic function in the standard form \( y = ax^2 + bx + c \), the vertex can be found using the formula

• \( h = -\frac{b}{2a} \) for the x-coordinate
• Substitute \( x = h \) back into the function to find the y-coordinate, \( k \)

Together, these give the vertex as \((h, k)\).
In our problem case, with \( a = 3 \) and \( b = 2 \), we calculated the vertex to be \((-\frac{1}{3}, \frac{25}{3})\). This means that the parabola's lowest point is at this vertex, as the parabola opens upwards.

Axis of Symmetry

The axis of symmetry is an imaginary vertical line that runs through the vertex of the parabola. It divides the parabola into two mirror-image halves. The equation for the axis of symmetry is derived directly from the x-coordinate of the vertex.
In simpler terms, the axis of symmetry for any quadratic function \( y = ax^2 + bx + c \) is given by the line \( x = h \), where \( h = -\frac{b}{2a} \).
For our specific quadratic equation, the axis of symmetry is \( x = -\frac{1}{3} \). This means that on a graph, all points on the left of this vertical line have corresponding points on the right, maintaining balance and symmetry in the parabola.

Quadratic Function Features

Understanding and identifying key features of quadratic functions helps in graphing them accurately without a calculator. Besides the vertex and axis of symmetry, other important features include:

• Y-intercept: This is where the parabola crosses the y-axis, occurring when \( x = 0 \). For our example, substituting \( x = 0 \) results in \( y = 4 \), giving us the point \( (0, 4) \).
• Local Minimum or Maximum: The vertex acts as either a minimum or a maximum point. If the parabola opens upwards (as in our case), the vertex is the minimum point— \((-\frac{1}{3}, \frac{25}{3})\).
• Direction of Opening: The sign of the coefficient \( a \) determines the direction. A positive \( a \) means the parabola opens upwards.

Each of these elements contributes to constructing the full shape and positioning of the parabola on a graph.

End Behavior of Quadratic Functions

The end behavior of quadratic functions describes what happens to the value of \( y \) as \( x \) approaches positive or negative infinity. By examining the leading term, \( 3x^2 \), of our quadratic \( y = 3x^2 + 2x + 4 \):

• As \( x \to +\infty \), \( y \to +\infty \)
• As \( x \to -\infty \), \( y \to +\infty \)

This is because \( 3x^2 \) dominates the behavior of the function, causing both ends of the graph to rise upwards. The graph extends upward indefinitely on both sides.
This characteristic is typical for all parabolas with a positive \( a \) coefficient, clearly visualizing a U-shape that just keeps going higher as it extends along the x-axis.