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The equation above represents a parabola in the xy-plane. If a < 0 and
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31 Mar 2021, 10:04
31 Mar 2021, 10:04
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\(f(x) = ax^2 + bx + c\)
The equation above represents a parabola in the xy-plane. If \(a < 0\) and the x-intercepts of the parabola are \(-1\) and \(3\), which of the following could be the vertex of the parabola?
A. \((2, 5)\)
B. \((1, -3)\)
C. \((2, 3)\)
D. \((1, 4)\)
E. \((2, -2)\)
The equation above represents a parabola in the xy-plane. If \(a < 0\) and the x-intercepts of the parabola are \(-1\) and \(3\), which of the following could be the vertex of the parabola?
A. \((2, 5)\)
B. \((1, -3)\)
C. \((2, 3)\)
D. \((1, 4)\)
E. \((2, -2)\)
Re: The equation above represents a parabola in the xy-plane. If a < 0 and
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31 Mar 2021, 20:20
31 Mar 2021, 20:20
2
The vertex of the parabola is basically the tip of the parabola. For the given parabola with a<0, the shape of the parabola is going to be inverted. The tip or the vertex must be at the top while open, diverging ends must be on the bottom.
Then the two x-intercepts are -1 and 3. Since a parabola is always symmetric through the vertex, the vertex must lie on the perpendicular line to x-axis between -1 and 3. BTW this line is also called axis of symmetry. So, the line must be x=1, falling exactly between the given two points. Vertex will be a point somewhere on this line. Let's eliminate the answer choices that do not have x=1.
Left with option B and D, we can predict that the vertex must be the one with the positive y-coordinate since the shape is inverted. If it were option B (1, -3), the tip would be below the x-axis, not giving any x-intercepts. Therefore, the vertex could be (1, 4).
D is the correct answer.
Then the two x-intercepts are -1 and 3. Since a parabola is always symmetric through the vertex, the vertex must lie on the perpendicular line to x-axis between -1 and 3. BTW this line is also called axis of symmetry. So, the line must be x=1, falling exactly between the given two points. Vertex will be a point somewhere on this line. Let's eliminate the answer choices that do not have x=1.
Left with option B and D, we can predict that the vertex must be the one with the positive y-coordinate since the shape is inverted. If it were option B (1, -3), the tip would be below the x-axis, not giving any x-intercepts. Therefore, the vertex could be (1, 4).
D is the correct answer.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: The equation above represents a parabola in the xy-plane. If a < 0 and
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02 Apr 2021, 06:51
02 Apr 2021, 06:51
1
KarunMendiratta wrote:
\(f(x) = ax^2 + bx + c\)
The equation above represents a parabola in the xy-plane. If \(a < 0\) and the x-intercepts of the parabola are \(-1\) and \(3\), which of the following could be the vertex of the parabola?
A. \((2, 5)\)
B. \((1, -3)\)
C. \((2, 3)\)
D. \((1, 4)\)
E. \((2, -2)\)
The equation above represents a parabola in the xy-plane. If \(a < 0\) and the x-intercepts of the parabola are \(-1\) and \(3\), which of the following could be the vertex of the parabola?
A. \((2, 5)\)
B. \((1, -3)\)
C. \((2, 3)\)
D. \((1, 4)\)
E. \((2, -2)\)
Explanation:
The vertex of Parabola lies at the mid-point of x-intercepts (or x values for which we have same y values)
Here, \(x_v = \frac{3 + (-1)}{2} = 1\)
Eliminate A, C, and E
Since, \(a < 0\), the Parabola will open downwards with the y-coordinate of vertex (Maximum point in this case) as a +ve value
Hence, option D
