Problem 14 - Calculus Single Variable

Question

Sketch the graph of the function defined by the given expression. $$ x^{2}+6 x+10 $$

Step-by-Step Solution

Answer

The graph is an upward-opening parabola with vertex \((-3, 1)\) and y-intercept at \((0, 10)\).

Step 1: Identify the Function Type

The given expression is a quadratic function, which generally has the form \( f(x) = ax^2 + bx + c \). In this case, \( a = 1 \), \( b = 6 \), and \( c = 10 \). Quadratic functions graph as parabolas.

Step 2: Determine the Direction of the Parabola

Since the coefficient of \( x^2 \) (\( a = 1 \)) is positive, the parabola opens upwards.

Step 3: Find the Vertex of the Parabola

For a quadratic function \( f(x) = ax^2 + bx + c \), the vertex \( (h, k) \) can be found using the formula \( h = -\frac{b}{2a} \). Substituting \( a = 1 \) and \( b = 6 \), we get \( h = -\frac{6}{2 \cdot 1} = -3 \). Substitute \( x = -3 \) into the function to find \( k \):\[ k = (-3)^2 + 6(-3) + 10 = 9 - 18 + 10 = 1. \]Thus, the vertex is \((-3, 1)\).

Step 4: Determine the Axis of Symmetry

The axis of symmetry for the parabola is the vertical line that passes through the vertex. Therefore, the axis of symmetry is \( x = -3 \).

Step 5: Find the Y-Intercept

The y-intercept is the value of the function when \( x = 0 \). Substituting \( x = 0 \) into the function, we find the y-intercept: \[ f(0) = 0^2 + 6 \cdot 0 + 10 = 10. \]So, the point is \((0, 10)\).

Step 6: Sketch the Graph

Plot the vertex \((-3, 1)\) and the y-intercept \((0, 10)\) on the coordinate plane. Since the parabola opens upwards, sketch a symmetrical curve about the axis of symmetry \( x = -3 \), ensuring it passes through the vertex and approaches the y-intercept. The graph is a smooth, symmetric U-shaped curve extending upwards.

Key Concepts

Graphing Parabolas

When graphing a quadratic function, like the one given in the expression \( x^2 + 6x + 10 \), you're essentially sketching what is known as a parabola. Parabolas are U-shaped curves that can either open upwards or downwards. The direction in which the parabola opens is determined by the coefficient of the \( x^2 \) term in the standard quadratic form \( ax^2 + bx + c \).

In our case, since \( a = 1 \) which is positive, the parabola opens upwards. Here are some tips to graph a parabola:

• Find key points like the vertex and the y-intercept.
• Use the axis of symmetry to ensure the parabola is symmetric.
• Make sure to include additional points as needed to accurately display the shape of the curve.

Graphing a parabola combines these elements to provide a visual representation of the quadratic function.

Vertex of a Parabola

The vertex of the parabola is a crucial point, representing the peak or the lowest point of the curve, depending on its orientation. For the function \( x^2 + 6x + 10 \), we need to find both coordinates of the vertex using specific formulas.

The x-coordinate of the vertex \( h \) can be calculated using the formula:
\[ h = -\frac{b}{2a} \]
Let’s substitute \( a = 1 \) and \( b = 6 \) into this formula:

• The calculation results in \( h = -3 \).

Next, substitute \( h = -3 \) back into the quadratic equation to find the y-coordinate \( k \), where:
\[ k = (-3)^2 + 6(-3) + 10 = 1 \]
The vertex of our parabola is the point \((-3, 1)\). This is the lowest point since our parabola opens upwards.

Axis of Symmetry

Every parabola has an axis of symmetry, which is a vertical line passing through the vertex, splitting the parabola into two mirror images. For our quadratic function, the axis of symmetry can be directly derived from the x-coordinate of the vertex.

Given that our vertex is at \((-3, 1)\), the axis of symmetry is located at:

• \( x = -3 \)

This line serves as a guide, ensuring that both halves of the parabola are evenly balanced. Drawing the axis of symmetry on the graph aids in making the parabola symmetric and balanced on either side of \( x = -3 \).

Y-Intercept

The y-intercept is the point where the graph of the function crosses the y-axis. It provides a convenient, easily located point during graphing. For any quadratic function, the y-intercept can be found by evaluating the function when \( x = 0 \).

In the case of our function \( x^2 + 6x + 10 \), substitute \( x = 0 \) into the equation:

• The calculation shows \( f(0) = 10 \).
• This means the y-intercept is the point \((0, 10)\).

Make sure to plot the y-intercept as a guide point on your graph, since it aids in sketching the overall curve and in ensuring that the graph accurately represents the function. It's one of the key features that define the path of the quadratic function as it extends.