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Sagot :
To graph the function [tex]\( f(x) = 8x^2 + 16x + 3 \)[/tex], follow these detailed steps:
1. Identify the function type:
- The function [tex]\( f(x) = 8x^2 + 16x + 3 \)[/tex] is a quadratic function, which means its graph will be a parabola.
2. Determine the general shape and opening direction:
- Since the coefficient of [tex]\( x^2 \)[/tex] is positive ([tex]\( 8 \)[/tex]), the parabola opens upwards.
3. Find the vertex:
- The vertex form of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] can be found by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Substituting [tex]\( a = 8 \)[/tex] and [tex]\( b = 16 \)[/tex]:
[tex]\[ x = -\frac{16}{2 \times 8} = -1 \][/tex]
- To find the y-coordinate of the vertex, substitute [tex]\( x = -1 \)[/tex] back into the function:
[tex]\[ f(-1) = 8(-1)^2 + 16(-1) + 3 = 8 - 16 + 3 = -5 \][/tex]
- Therefore, the vertex of the parabola is at [tex]\((-1, -5)\)[/tex].
4. Find the y-intercept:
- Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = 8(0)^2 + 16(0) + 3 = 3 \][/tex]
- Thus, the y-intercept is at [tex]\((0, 3)\)[/tex].
5. Plotting points:
- To generate points for a smooth graph, we can calculate [tex]\( f(x) \)[/tex] for a range of [tex]\( x \)[/tex]-values. In our case, we've already calculated a detailed list of points:
- x-values:
[tex]\[ \{-10, -9.94987469, -9.89974937, \ldots, 9.94987469, 10\} \][/tex]
- Corresponding y-values:
[tex]\[ \{643, 635.802055, 628.644311, \ldots, 943.36291, 960\} \][/tex]
6. Plotting the points and graph:
- First, plot the vertex [tex]\((-1, -5)\)[/tex] and the y-intercept [tex]\((0, 3)\)[/tex].
- Then, mark additional points derived from our calculated values:
[tex]\[ \{(-10, 643), (-9, 472), \ldots, (9, 945), (10, 960)\} \][/tex]
By plotting several points along with the vertex and the y-intercept, you can sketch the parabola.
7. Drawing the parabola:
- Connect the plotted points smoothly to form the upwards-opening parabola. The symmetry will be around the vertical line [tex]\(x = -1\)[/tex].
By following these steps, you can graph the quadratic function [tex]\( f(x) = 8x^2 + 16x + 3 \)[/tex], ensuring that the vertex, y-intercept, and a selection of calculated points are accurately represented to create a smooth curve.
1. Identify the function type:
- The function [tex]\( f(x) = 8x^2 + 16x + 3 \)[/tex] is a quadratic function, which means its graph will be a parabola.
2. Determine the general shape and opening direction:
- Since the coefficient of [tex]\( x^2 \)[/tex] is positive ([tex]\( 8 \)[/tex]), the parabola opens upwards.
3. Find the vertex:
- The vertex form of a quadratic function [tex]\( ax^2 + bx + c \)[/tex] can be found by:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Substituting [tex]\( a = 8 \)[/tex] and [tex]\( b = 16 \)[/tex]:
[tex]\[ x = -\frac{16}{2 \times 8} = -1 \][/tex]
- To find the y-coordinate of the vertex, substitute [tex]\( x = -1 \)[/tex] back into the function:
[tex]\[ f(-1) = 8(-1)^2 + 16(-1) + 3 = 8 - 16 + 3 = -5 \][/tex]
- Therefore, the vertex of the parabola is at [tex]\((-1, -5)\)[/tex].
4. Find the y-intercept:
- Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = 8(0)^2 + 16(0) + 3 = 3 \][/tex]
- Thus, the y-intercept is at [tex]\((0, 3)\)[/tex].
5. Plotting points:
- To generate points for a smooth graph, we can calculate [tex]\( f(x) \)[/tex] for a range of [tex]\( x \)[/tex]-values. In our case, we've already calculated a detailed list of points:
- x-values:
[tex]\[ \{-10, -9.94987469, -9.89974937, \ldots, 9.94987469, 10\} \][/tex]
- Corresponding y-values:
[tex]\[ \{643, 635.802055, 628.644311, \ldots, 943.36291, 960\} \][/tex]
6. Plotting the points and graph:
- First, plot the vertex [tex]\((-1, -5)\)[/tex] and the y-intercept [tex]\((0, 3)\)[/tex].
- Then, mark additional points derived from our calculated values:
[tex]\[ \{(-10, 643), (-9, 472), \ldots, (9, 945), (10, 960)\} \][/tex]
By plotting several points along with the vertex and the y-intercept, you can sketch the parabola.
7. Drawing the parabola:
- Connect the plotted points smoothly to form the upwards-opening parabola. The symmetry will be around the vertical line [tex]\(x = -1\)[/tex].
By following these steps, you can graph the quadratic function [tex]\( f(x) = 8x^2 + 16x + 3 \)[/tex], ensuring that the vertex, y-intercept, and a selection of calculated points are accurately represented to create a smooth curve.
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