C). y = (x − 3)² − 2
We will solve the problems associated with the quadratic functions.
Given:
A graph shows a parabola that opens up and crosses the x-axis near one and a half and four and a half.
Question:
What is the equation of the graph below?
A) y = − (x − 3)² − 2
B) y = − (x + 2)² − 3
C) y = (x − 3)² − 2
D) y = (x + 2)² − 3
The Process:
If we know the vertex point of the parabola, V(h, k), then we use the formula: [tex]\boxed{\boxed{ \ y = a(x - h)^2 + k \ }}[/tex]
When [tex]\boxed{ \ a > 0 \ }[/tex], the parabola opens upward. Therefore we have C or D as a candidate for the answer.
The axis of symmetry is the line that passes through the vertex of a parabola with equation: [tex]\boxed{ \ x = h = - \frac{b}{2a} \ }[/tex]
If (x₁, 0) and (x₂,0) are the x-intercepts of the graph, then the axis of symmetry can also be determined by [tex]\boxed{ \ x = h = \frac{x_1 + x_2}{2} \ }[/tex]
From the problem, the graph crosses the x-axis near one and a half and four and a half. So we can find the axis of the symmetry.
[tex]\boxed{ \ x = h = \frac{1.5 + 4.5}{2} \ }[/tex]
[tex]\boxed{ \ h = 3 \ }[/tex]
Thus, we have found the equation of the graph, which is [tex]\boxed{\boxed{ \ C) \ y = (x - 3)^2 - 2 \ }}[/tex].
From the equation of the graph, we know that the vertex point of the parabola is (h, k) = (3, -2), resulting in a vertex that is a minimum.
See the attachment.
- - - - - - - - - -
Notes:
Let us change to standard equation form.
y = (x − 3)² − 2
y = x² - 6x + 9 - 2
y = x² - 6x + 7
A quadratic function is described by the standard equation [tex]\boxed{\boxed{ \ f(x) = ax^2 + bx + c \ }}[/tex], and the value c is the y-intercept of the graph.
See the attachment once again, the y-intercept is (0, 7).
Keywords: a graph, shows a parabola, that opens up, crosses the x-axis, near one and a half, four, the quadratic functions, the equation of the graph, vertex, the axis of symmetry, opens upward, the x-intercept, the y-intercept, minimum
The equation of the graph is [tex]\boxed{y = {{\left( {x - 3} \right)}^2} - 2}.[/tex]Option (c) is correct.
Further explanation:
The general equation of the parabola can be expressed as follows,
[tex]\boxed{y = a{{\left( {x - h} \right)}^2} + k}[/tex]
Here, h and k the vertices.
Given:
The equations are as follows,
(a). [tex]y = - {\left( {x - 3} \right)^2} - 2[/tex]
(b). [tex]y = - {\left( {x + 2} \right)^2} - 3[/tex]
(c). [tex]y = {\left( {x - 3} \right)^2} - 2[/tex]
(d). [tex]y = {\left( {x + 2} \right)^2} - 3[/tex]
Explanation:
If the value of a is greater than zero than the parabola is an upward.
If the value of a is less than zero than the parabola is downward.
The value of a in equation (a) and equation (b) is less than zero. Therefore, option (a) and option (b) is not correct.
The x-intercepts are 1.5 and 4.5.
The value of h can be obtained as follows,
[tex]\begin{aligned}h&= \frac{{1.5 + 4.5}}{2}\\&= \frac{6}{2} \\ &= 3\\\end{aligned}[/tex]
Hence, the equation is [tex]y = {\left( {x - 3} \right)^2} - 2.[/tex]
The equation of the graph is [tex]\boxed{y = {{\left( {x - 3} \right)}^2} - 2}[/tex]. Option (c) is correct.
Option (a) is not correct.
Option (b) is not correct.
Option (c) is correct.
Option (d) is not correct.
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Polynomials
Keywords: parabola, graph, shows, opens up, crosses x-axis, one, one and a half, and four and a half, quadratic equation, equation factorization. Factorized form, polynomial, quadratic formula, zeroes, Fundamental Theorem of algebra, polynomial
