Respuesta :
Answer:
Step-by-step explanation:
A. f(x) = (2x-3)(x-1)
B. since the coefficient of x^2 is positive, the parabola opens upward.
C. plot the vetex, which is between the roots, at x = 5/4, plot the zeroes
plot the points at x=1,2, draw a curve through them.
You can use any of several handy online graphing tools to verify your work.
Solving the quadratic equation, and finding it's vertex, it is found that:
- a) The x-intercepts are [tex]x = \frac{3}{2}[/tex] and [tex]x = 1[/tex].
- b) Since the coefficient a is positive, it is a minimum. The coordinates of the vertex are [tex](\frac{5}{4}, -\frac{1}{8})[/tex]
- c) Using the x-intercepts, the vertex and the y-intercept, the graph is given at the end of this answer.
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Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots, also called x-intercepts, [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}[/tex]
[tex]\Delta = b^{2} - 4ac[/tex]
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Vertex of a quadratic function:
Suppose we have a quadratic function in the following format:
[tex]f(x) = ax^{2} + bx + c[/tex]
It's vertex is the point [tex](x_{v}, y_{v})[/tex]
In which
[tex]x_{v} = -\frac{b}{2a}[/tex]
[tex]y_{v} = -\frac{\Delta}{4a}[/tex]
Where
[tex]\Delta = b^2-4ac[/tex]
If a<0, the vertex is a maximum point, that is, the maximum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
If a>0, the vertex is a minimum point, that is, the minimum value happens at [tex]x_{v}[/tex], and it's value is [tex]y_{v}[/tex].
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Question a:
We have the following quadratic equation:
[tex]f(x) = 2x^2 - 5x + 3[/tex]
Which has [tex]a = 2, b = -5, c = 3[/tex]
The x-intercepts are the roots, so:
[tex]\Delta = (-5)^2 - 4(2)(3) = 1[/tex]
[tex]x_{1} = \frac{-(-5) + \sqrt{1}}{2(2)} = \frac{3}{2}[/tex]
[tex]x_{2} = \frac{-(-5) - \sqrt{1}}{2(2)} = 1[/tex]
The x-intercepts are [tex]x = \frac{3}{2}[/tex] and [tex]x = 1[/tex].
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Question b:
Since the coefficient a is positive, it is a minimum.
To find the coordinates, we apply the formula. Thus:
[tex]x_{v} = -\frac{(-5)}{2(2)} = \frac{5}{4}[/tex]
[tex]y_{v} = -\frac{1}{4(2)} = -\frac{1}{8}[/tex]
The coordinates of the vertex are [tex](\frac{5}{4}, -\frac{1}{8})[/tex]
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Question c:
- The x-intercepts are [tex]x = \frac{3}{2}[/tex] and [tex]x = 1[/tex].
- The coordinates of the vertex are [tex](\frac{5}{4}, -\frac{1}{8})[/tex]
- The y-intercept is at [tex]y = f(0) = 3[/tex].
Using this, we get the graph given at the end of this answer.
A similar question is given at https://brainly.com/question/22851074
