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Magazine Chapter 1: Problem 14
Sketch the graph of the function defined by the given expression. $$ x^{2}+6 x+10 $$
Short Answer
Expert verified
The graph is an upward-opening parabola with vertex \((-3, 1)\) and y-intercept at \((0, 10)\).
Step by step solution
01
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.
02
Determine the Direction of the Parabola
Since the coefficient of \( x^2 \) (\( a = 1 \)) is positive, the parabola opens upwards.
03
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 \((-3, 1)\).
04
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 \( x = -3 \).
05
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 \((0, 10)\).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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.
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鈥檚 substitute \( a = 1 \) and \( b = 6 \) into this formula:
\[ k = (-3)^2 + 6(-3) + 10 = 1 \]
The vertex of our parabola is the point \((-3, 1)\). This is the lowest point since our parabola opens upwards.
The x-coordinate of the vertex \( h \) can be calculated using the formula:
\[ h = -\frac{b}{2a} \]
Let鈥檚 substitute \( a = 1 \) and \( b = 6 \) into this formula:
- The calculation results in \( h = -3 \).
\[ k = (-3)^2 + 6(-3) + 10 = 1 \]
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:
Given that our vertex is at \((-3, 1)\), the axis of symmetry is located at:
- \( x = -3 \)
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:
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 \((0, 10)\).
