You should notice that the domain and the range of quadratic functio.pdf
You should notice that the domain and the range of quadratic function [ f(x) = x^2 - 4] are the same set of real numbers, R. Since the given function is quadratic, its graph is a parabola. You may find where the parabola intercepts x axis, considering [f(x) = 0] , such that: [x^2 - 4 = 0 => x^2 = 4 => x_1 = -2; x_2 = 2] Hence, parabola intercepts x a xis at [(-2,0)] and [(2,0).] You also may find the vertex of parabola, using the following formulas, such that: [x = -b/(2a); y = (4ac-b^2)/(4a)] Since [a = 1, b = 0, c = -4] , yields [x = -0/2 => x = 0] [y = (-16 - 0)/(4) => y = -4] Hence, evaluating the vertex of parabola yields [V(0,-4).] Thus, you may find important points on parabola not only sketching the graph, but using the formulas indicate above. Sketching the parabola, you may notice the points of interception and the position of vertex. Solution You should notice that the domain and the range of quadratic function [ f(x) = x^2 - 4] are the same set of real numbers, R. Since the given function is quadratic, its graph is a parabola. You may find where the parabola intercepts x axis, considering [f(x) = 0] , such that: [x^2 - 4 = 0 => x^2 = 4 => x_1 = -2; x_2 = 2] Hence, parabola intercepts x a xis at [(-2,0)] and [(2,0).] You also may find the vertex of parabola, using the following formulas, such that: [x = -b/(2a); y = (4ac-b^2)/(4a)] Since [a = 1, b = 0, c = -4] , yields [x = -0/2 => x = 0] [y = (-16 - 0)/(4) => y = -4] Hence, evaluating the vertex of parabola yields [V(0,-4).] Thus, you may find important points on parabola not only sketching the graph, but using the formulas indicate above. Sketching the parabola, you may notice the points of interception and the position of vertex..
You should notice that the domain and the range of quadratic functio.pdf
You should notice that the domain and the range of quadratic function [ f(x) = x^2 - 4] are the same set of real numbers, R. Since the given function is quadratic, its graph is a parabola. You may find where the parabola intercepts x axis, considering [f(x) = 0] , such that: [x^2 - 4 = 0 => x^2 = 4 => x_1 = -2; x_2 = 2] Hence, parabola intercepts x a xis at [(-2,0)] and [(2,0).] You also may find the vertex of parabola, using the following formulas, such that: [x = -b/(2a); y = (4ac-b^2)/(4a)] Since [a = 1, b = 0, c = -4] , yields [x = -0/2 => x = 0] [y = (-16 - 0)/(4) => y = -4] Hence, evaluating the vertex of parabola yields [V(0,-4).] Thus, you may find important points on parabola not only sketching the graph, but using the formulas indicate above. Sketching the parabola, you may notice the points of interception and the position of vertex. Solution You should notice that the domain and the range of quadratic function [ f(x) = x^2 - 4] are the same set of real numbers, R. Since the given function is quadratic, its graph is a parabola. You may find where the parabola intercepts x axis, considering [f(x) = 0] , such that: [x^2 - 4 = 0 => x^2 = 4 => x_1 = -2; x_2 = 2] Hence, parabola intercepts x a xis at [(-2,0)] and [(2,0).] You also may find the vertex of parabola, using the following formulas, such that: [x = -b/(2a); y = (4ac-b^2)/(4a)] Since [a = 1, b = 0, c = -4] , yields [x = -0/2 => x = 0] [y = (-16 - 0)/(4) => y = -4] Hence, evaluating the vertex of parabola yields [V(0,-4).] Thus, you may find important points on parabola not only sketching the graph, but using the formulas indicate above. Sketching the parabola, you may notice the points of interception and the position of vertex..
![You should notice that the domain and the range of quadratic function [ f(x) = x^2 - 4] are the
same set of real numbers, R.
Since the given function is quadratic, its graph is a parabola. You may find where the parabola
intercepts x axis, considering [f(x) = 0] , such that:
[x^2 - 4 = 0 => x^2 = 4 => x_1 = -2; x_2 = 2]
Hence, parabola intercepts x a xis at [(-2,0)] and [(2,0).]
You also may find the vertex of parabola, using the following formulas, such that:
[x = -b/(2a); y = (4ac-b^2)/(4a)]
Since [a = 1, b = 0, c = -4] , yields
[x = -0/2 => x = 0]
[y = (-16 - 0)/(4) => y = -4]
Hence, evaluating the vertex of parabola yields [V(0,-4).]
Thus, you may find important points on parabola not only sketching the graph, but using the
formulas indicate above.
Sketching the parabola, you may notice the points of interception and the position of vertex.
Solution
You should notice that the domain and the range of quadratic function [ f(x) = x^2 - 4] are the
same set of real numbers, R.
Since the given function is quadratic, its graph is a parabola. You may find where the parabola
intercepts x axis, considering [f(x) = 0] , such that:
[x^2 - 4 = 0 => x^2 = 4 => x_1 = -2; x_2 = 2]
Hence, parabola intercepts x a xis at [(-2,0)] and [(2,0).]
You also may find the vertex of parabola, using the following formulas, such that:
[x = -b/(2a); y = (4ac-b^2)/(4a)]
Since [a = 1, b = 0, c = -4] , yields
[x = -0/2 => x = 0]
[y = (-16 - 0)/(4) => y = -4]
Hence, evaluating the vertex of parabola yields [V(0,-4).]
Thus, you may find important points on parabola not only sketching the graph, but using the
formulas indicate above.
Sketching the parabola, you may notice the points of interception and the position of vertex.](https://image.slidesharecdn.com/youshouldnoticethatthedomainandtherangeofquadraticfunctio-230410075638-f074c1df/75/you-should-notice-that-the-domain-and-the-range-of-quadratic-functiopdf-1-2048.jpg?cb=1681115835)
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You should notice that the domain and the range of quadratic functio.pdf
- 1. You should notice that the domain and the range of quadratic function [ f(x) = x^2 - 4] are the same set of real numbers, R. Since the given function is quadratic, its graph is a parabola. You may find where the parabola intercepts x axis, considering [f(x) = 0] , such that: [x^2 - 4 = 0 => x^2 = 4 => x_1 = -2; x_2 = 2] Hence, parabola intercepts x a xis at [(-2,0)] and [(2,0).] You also may find the vertex of parabola, using the following formulas, such that: [x = -b/(2a); y = (4ac-b^2)/(4a)] Since [a = 1, b = 0, c = -4] , yields [x = -0/2 => x = 0] [y = (-16 - 0)/(4) => y = -4] Hence, evaluating the vertex of parabola yields [V(0,-4).] Thus, you may find important points on parabola not only sketching the graph, but using the formulas indicate above. Sketching the parabola, you may notice the points of interception and the position of vertex. Solution You should notice that the domain and the range of quadratic function [ f(x) = x^2 - 4] are the same set of real numbers, R. Since the given function is quadratic, its graph is a parabola. You may find where the parabola intercepts x axis, considering [f(x) = 0] , such that: [x^2 - 4 = 0 => x^2 = 4 => x_1 = -2; x_2 = 2] Hence, parabola intercepts x a xis at [(-2,0)] and [(2,0).] You also may find the vertex of parabola, using the following formulas, such that: [x = -b/(2a); y = (4ac-b^2)/(4a)] Since [a = 1, b = 0, c = -4] , yields [x = -0/2 => x = 0] [y = (-16 - 0)/(4) => y = -4] Hence, evaluating the vertex of parabola yields [V(0,-4).] Thus, you may find important points on parabola not only sketching the graph, but using the formulas indicate above. Sketching the parabola, you may notice the points of interception and the position of vertex.