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If the curve represented by y = x^2 – 5x + t intersects with
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22 Apr 2020, 12:35
22 Apr 2020, 12:35
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If the curve represented by \(y = x^2 – 5x + t\) intersects with the x-axis at two points and one of the points is (–1, 0), what is the other point?
(A) (1, 0)
(B) (–2, 0)
(C) (5, 0)
(D) (6, 0)
(E) (3, 0)
(A) (1, 0)
(B) (–2, 0)
(C) (5, 0)
(D) (6, 0)
(E) (3, 0)
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Re: If the curve represented by y = x^2 – 5x + t intersects with
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22 Apr 2020, 12:55
22 Apr 2020, 12:55
Carcass wrote:
If the curve represented by \(y = x^2 – 5x + t\) intersects with the x-axis at two points and one of the points is (–1, 0), what is the other point?
(A) (1, 0)
(B) (–2, 0)
(C) (5, 0)
(D) (6, 0)
(E) (3, 0)
(A) (1, 0)
(B) (–2, 0)
(C) (5, 0)
(D) (6, 0)
(E) (3, 0)
If the point (-1, 0) is ON the line, then x = -1 and y = 0 is a SOLUTION to the curve's equation y = x² - 5x + t
That is: 0 = (-1)² - 5(-1) + t
Evaluate: 0 = 1 + 5 + t
Solve: t = -6
So, the original equation is y = x² - 5x + (-6), which is the same as y = x² - 5x - 6
Our goal is to find the OTHER point that intersects with the x-axis
IMPORTANT: The y-coordinate of ANY point on the x-axis is 0
So, the coordinates of the OTHER point of intersection is: (x, 0)
Our job is to determine the value of x
To do so, plug y = 0 into the equation to get: 0 = x² - 5x - 6
Factor the right side: 0 = (x + 1)(x - 6)
So, the solutions are x = -1 and x = 6
x = -1 is already noted in the point of intersection (-1, 0)
So, the OTHER point of intersection (with the x-axis) is (6, 0)
Answer: D
Cheers,
Brent
Re: If the curve represented by y = x^2 – 5x + t intersects with
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10 May 2020, 13:10
10 May 2020, 13:10
Carcass wrote:
If the curve represented by \(y = x^2 – 5x + t\) intersects with the x-axis at two points and one of the points is (–1, 0), what is the other point?
(A) (1, 0)
(B) (–2, 0)
(C) (5, 0)
(D) (6, 0)
(E) (3, 0)
(A) (1, 0)
(B) (–2, 0)
(C) (5, 0)
(D) (6, 0)
(E) (3, 0)
Another approach uses the fact that x-coordinate of the parabola's vertex = -b/2a, where a and b are from the equation in standard form: ax^2+bx+c.
In this case, a=1, b=-5. Therefore, the parabola's vertex has x-coordinate = 5/2 or 2.5.
Imagine the parabola visually. It opens upward and crosses the x axis at two points (one of which we already know). Each x-intercept is equidistant from the x-coordinate of the vertex.
Since we know that one x-int is at x=-1, which is 3.5 units away from the vertex at x=2.5, we can reason that the other x-int, to the right of the vertex, will have an x-coordinate that is equal to 2.5+3.5, or 6.
