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y = α(x − 2)(x + 4)
In the quadratic equation above, a is a nonzero constant. The graph of the equation in the xy-plane is a parabola with vertex (c, d). Which of the following is equal to d?
In the quadratic equation above, a is a nonzero constant. The graph of the equation in the xy-plane is a parabola with vertex (c, d). Which of the following is equal to d?
- a)−9α
- b)−8α
- c)−5α
- d)−2α
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
y = α(x − 2)(x + 4)In the quadratic equation above, a is a...
Choice A is correct. The parabola with equation y = α(x − 2)(x + 4) crosses the x-axis at the points (−4, 0) and (2, 0). The x-coordinate of the vertex of the parabola is halfway between the x-coordinates of (−4, 0) and (2, 0). Thus, the x-coordinate of the vertex is 
= −1. This is the value of c. To find the y-coordinate of the vertex, substitute −1 for x in y = α(x − 2)(x + 4): y = α(x − 2)(x + 4) = α(−1 − 2)(−1 + 4) = α(−3)(3) = −9α.
Therefore, the value of d is −9α.
Choice B is incorrect because the value of the constant term in the equation is not the y-coordinate of the vertex, unless there were no linear terms in the quadratic. Choice C is incorrect and may be the result of a sign error in finding the x-coordinate of the vertex. Choice D is incorrect because the negative of the coefficient of the linear term in the quadratic is not the y-coordinate of the vertex.
= −1. This is the value of c. To find the y-coordinate of the vertex, substitute −1 for x in y = α(x − 2)(x + 4): y = α(x − 2)(x + 4) = α(−1 − 2)(−1 + 4) = α(−3)(3) = −9α.Therefore, the value of d is −9α.
Choice B is incorrect because the value of the constant term in the equation is not the y-coordinate of the vertex, unless there were no linear terms in the quadratic. Choice C is incorrect and may be the result of a sign error in finding the x-coordinate of the vertex. Choice D is incorrect because the negative of the coefficient of the linear term in the quadratic is not the y-coordinate of the vertex.
Most Upvoted Answer
y = α(x − 2)(x + 4)In the quadratic equation above, a is a...
Choice A is correct. The parabola with equation y = α(x − 2)(x + 4) crosses the x-axis at the points (−4, 0) and (2, 0). The x-coordinate of the vertex of the parabola is halfway between the x-coordinates of (−4, 0) and (2, 0). Thus, the x-coordinate of the vertex is = −1. This is the value of c. To find the y-coordinate of the vertex, substitute −1 for x in y = α(x − 2)(x + 4): y = α(x − 2)(x + 4) = α(−1 − 2)(−1 + 4) = α(−3)(3) = −9α.
Therefore, the value of d is −9α.
Choice B is incorrect because the value of the constant term in the equation is not the y-coordinate of the vertex, unless there were no linear terms in the quadratic. Choice C is incorrect and may be the result of a sign error in finding the x-coordinate of the vertex. Choice D is incorrect because the negative of the coefficient of the linear term in the quadratic is not the y-coordinate of the vertex.
= −1. This is the value of c. To find the y-coordinate of the vertex, substitute −1 for x in y = α(x − 2)(x + 4): y = α(x − 2)(x + 4) = α(−1 − 2)(−1 + 4) = α(−3)(3) = −9α.Therefore, the value of d is −9α.
Choice B is incorrect because the value of the constant term in the equation is not the y-coordinate of the vertex, unless there were no linear terms in the quadratic. Choice C is incorrect and may be the result of a sign error in finding the x-coordinate of the vertex. Choice D is incorrect because the negative of the coefficient of the linear term in the quadratic is not the y-coordinate of the vertex.
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y = α(x − 2)(x + 4)In the quadratic equation above, a is a...
Understanding the Quadratic Equation
The equation given is y = α(x - 2)(x + 4). This is a quadratic equation in factored form, where α is a nonzero constant.
Finding the Vertex
To find the vertex of the quadratic, we can rewrite the equation in standard form by expanding it:
- Expand: y = α[(x - 2)(x + 4)] = α(x^2 + 2x - 8) = αx^2 + 2αx - 8α.
The vertex form of a quadratic equation is given by y = a(x - h)^2 + k, where (h, k) is the vertex.
Finding the Vertex Coordinates
For a quadratic in standard form y = ax^2 + bx + c, the x-coordinate of the vertex (h) can be found using the formula:
- h = -b/(2a).
In our case:
- a = α
- b = 2α
Thus,
- h = -2α/(2α) = -1.
Now, substitute h back into the equation to find the y-coordinate (d):
- d = α(-1 - 2)(-1 + 4) = α(-3)(3) = -9α.
Conclusion
Therefore, the y-coordinate of the vertex (d) is equal to -9α. This matches option A (-9α), confirming that it is indeed the correct answer.
The equation given is y = α(x - 2)(x + 4). This is a quadratic equation in factored form, where α is a nonzero constant.
Finding the Vertex
To find the vertex of the quadratic, we can rewrite the equation in standard form by expanding it:
- Expand: y = α[(x - 2)(x + 4)] = α(x^2 + 2x - 8) = αx^2 + 2αx - 8α.
The vertex form of a quadratic equation is given by y = a(x - h)^2 + k, where (h, k) is the vertex.
Finding the Vertex Coordinates
For a quadratic in standard form y = ax^2 + bx + c, the x-coordinate of the vertex (h) can be found using the formula:
- h = -b/(2a).
In our case:
- a = α
- b = 2α
Thus,
- h = -2α/(2α) = -1.
Now, substitute h back into the equation to find the y-coordinate (d):
- d = α(-1 - 2)(-1 + 4) = α(-3)(3) = -9α.
Conclusion
Therefore, the y-coordinate of the vertex (d) is equal to -9α. This matches option A (-9α), confirming that it is indeed the correct answer.
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y = α(x − 2)(x + 4)In the quadratic equation above, a is a nonzero constant. The graph of the equation in the xy-plane is a parabola with vertex (c, d). Which of the following is equal to d?a)−9αb)−8αc)−5αd)−2αCorrect answer is option 'A'. Can you explain this answer?
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