Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
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 revisit us for more reliable answers to any questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.