Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
To sketch the graph of the quadratic function [tex]\( y = x^2 - 7x + 12 \)[/tex], we need to follow a series of steps, including finding the key features of the graph such as the roots (the x-intercepts), the vertex, and the y-intercept. Let's break this down step-by-step:
1. Determine the roots (x-intercepts) of the function:
To find the roots of the quadratic equation, we solve [tex]\( x^2 - 7x + 12 = 0 \)[/tex].
The quadratic can be factored as:
[tex]\[ x^2 - 7x + 12 = (x - 3)(x - 4) \][/tex]
Setting each factor equal to zero gives:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x - 4 = 0 \][/tex]
Thus, the roots are:
[tex]\[ x = 3 \quad \text{and} \quad x = 4 \][/tex]
2. Determine the vertex of the parabola:
The vertex form of a quadratic function [tex]\( y = ax^2 + bx + c \)[/tex] can be used to find the vertex. The x-coordinate of the vertex is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For [tex]\( y = x^2 - 7x + 12 \)[/tex], [tex]\( a = 1 \)[/tex] and [tex]\( b = -7 \)[/tex]. Thus,
[tex]\[ x = -\frac{-7}{2 \times 1} = \frac{7}{2} = 3.5 \][/tex]
To find the y-coordinate of the vertex, substitute [tex]\( x = 3.5 \)[/tex] into the equation:
[tex]\[ y = (3.5)^2 - 7(3.5) + 12 \][/tex]
Calculate each term separately:
[tex]\[ (3.5)^2 = 12.25, \quad -7(3.5) = -24.5, \quad \text{and} \quad y = 12.25 - 24.5 + 12 \][/tex]
[tex]\[ y = 12.25 - 24.5 + 12 = -0.25 \][/tex]
Thus, the vertex is at:
[tex]\[ (3.5, -0.25) \][/tex]
3. Determine the y-intercept:
The y-intercept is where the graph crosses the y-axis, which occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 0^2 - 7(0) + 12 = 12 \][/tex]
Therefore, the y-intercept is at:
[tex]\[ (0, 12) \][/tex]
4. Sketch the graph:
- Plot the roots (x-intercepts) at [tex]\( (3, 0) \)[/tex] and [tex]\( (4, 0) \)[/tex].
- Plot the vertex at [tex]\( (3.5, -0.25) \)[/tex].
- Plot the y-intercept at [tex]\( (0, 12) \)[/tex].
- Draw a smooth curve to represent the parabola, connecting these points.
Since the coefficient of [tex]\( x^2 \)[/tex] is positive ([tex]\( a = 1 \)[/tex]), the parabola opens upwards.
### Final Sketch
Here is the rough sketch of the graph of [tex]\( y = x^2 - 7x + 12 \)[/tex]:
- The vertex is at [tex]\( (3.5, -0.25) \)[/tex].
- The roots are at [tex]\( (3, 0) \)[/tex] and [tex]\( (4, 0) \)[/tex].
- The y-intercept is at [tex]\( (0, 12) \)[/tex].
[tex]\[ \begin{array}{l} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad y\\ \quad 12\_/\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad\_/\quad \cup \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \,1\_/\quad\quad\_/\quad \\ (0,12)\quad \quad(3,0)\quad(4,0)\\ \quad \quad \quad \quad \quad(3.5, -0.25)\\ \quad \quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad\quad\quad\quad\quad\quad\quad x\\ \end{array} \][/tex]
1. Determine the roots (x-intercepts) of the function:
To find the roots of the quadratic equation, we solve [tex]\( x^2 - 7x + 12 = 0 \)[/tex].
The quadratic can be factored as:
[tex]\[ x^2 - 7x + 12 = (x - 3)(x - 4) \][/tex]
Setting each factor equal to zero gives:
[tex]\[ x - 3 = 0 \quad \text{or} \quad x - 4 = 0 \][/tex]
Thus, the roots are:
[tex]\[ x = 3 \quad \text{and} \quad x = 4 \][/tex]
2. Determine the vertex of the parabola:
The vertex form of a quadratic function [tex]\( y = ax^2 + bx + c \)[/tex] can be used to find the vertex. The x-coordinate of the vertex is given by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For [tex]\( y = x^2 - 7x + 12 \)[/tex], [tex]\( a = 1 \)[/tex] and [tex]\( b = -7 \)[/tex]. Thus,
[tex]\[ x = -\frac{-7}{2 \times 1} = \frac{7}{2} = 3.5 \][/tex]
To find the y-coordinate of the vertex, substitute [tex]\( x = 3.5 \)[/tex] into the equation:
[tex]\[ y = (3.5)^2 - 7(3.5) + 12 \][/tex]
Calculate each term separately:
[tex]\[ (3.5)^2 = 12.25, \quad -7(3.5) = -24.5, \quad \text{and} \quad y = 12.25 - 24.5 + 12 \][/tex]
[tex]\[ y = 12.25 - 24.5 + 12 = -0.25 \][/tex]
Thus, the vertex is at:
[tex]\[ (3.5, -0.25) \][/tex]
3. Determine the y-intercept:
The y-intercept is where the graph crosses the y-axis, which occurs when [tex]\( x = 0 \)[/tex]. Substitute [tex]\( x = 0 \)[/tex] into the equation:
[tex]\[ y = 0^2 - 7(0) + 12 = 12 \][/tex]
Therefore, the y-intercept is at:
[tex]\[ (0, 12) \][/tex]
4. Sketch the graph:
- Plot the roots (x-intercepts) at [tex]\( (3, 0) \)[/tex] and [tex]\( (4, 0) \)[/tex].
- Plot the vertex at [tex]\( (3.5, -0.25) \)[/tex].
- Plot the y-intercept at [tex]\( (0, 12) \)[/tex].
- Draw a smooth curve to represent the parabola, connecting these points.
Since the coefficient of [tex]\( x^2 \)[/tex] is positive ([tex]\( a = 1 \)[/tex]), the parabola opens upwards.
### Final Sketch
Here is the rough sketch of the graph of [tex]\( y = x^2 - 7x + 12 \)[/tex]:
- The vertex is at [tex]\( (3.5, -0.25) \)[/tex].
- The roots are at [tex]\( (3, 0) \)[/tex] and [tex]\( (4, 0) \)[/tex].
- The y-intercept is at [tex]\( (0, 12) \)[/tex].
[tex]\[ \begin{array}{l} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad y\\ \quad 12\_/\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad\_/\quad \cup \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \,1\_/\quad\quad\_/\quad \\ (0,12)\quad \quad(3,0)\quad(4,0)\\ \quad \quad \quad \quad \quad(3.5, -0.25)\\ \quad \quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad\quad\quad\quad\quad\quad\quad x\\ \end{array} \][/tex]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.