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If the curve represented by y = x^2 – 5x + t intersects ..

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by VJesus12 » Sat Jan 27, 2018 8:07 am
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)

The OA is the option D.

How can I find the second point without knowing the value of t? Experts, can you give me some help? Thanks.
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by Brent@GMATPrepNow » Sat Jan 27, 2018 8:20 am
VJesus12 wrote:If the curve represented by y = x² - 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)
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
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by EconomistGMATTutor » Sat Jan 27, 2018 8:34 am
Hi Vjesus12.

Let's take a look at your question.

Option 1:

Since $$y=x^2-5x+t$$ intersects with the x-axis at (-1,0) we have that $$0=\left(-1\right)^2-5\left(-1\right)+t\ \Leftrightarrow\ 0=1+5+t\ \Leftrightarrow\ t=-6.$$ Hence, the curve is represented by $$y=x^2-5x-6,$$ and this cuadratic equation has to roots, $$x_1=-1\ \left(\text{already}\ \text{known}\right)\ and\ \ \ \ \ x_2=6.$$ Therefore, the curve intersects with the x-axis at (6,0).

This is why the correct answer is D.

Option 2:

Since $$y=x^2-5x+t$$ we have that $$\ x^2-5x+t=\left(x-x_1\right)\left(x-x_2\right).$$ We already know that x_1=-1. Therefore, $$x^2-5x+t=\left(x+1\right)\left(x-x_2\right)\ and\ \ \left(\ 1-x_2=-5\ and\ \ 1\cdot (-x_2)=-x_2=t\right)$$ and therefore $$x_2=6.$$ Therefore, the curve intersects with the x-axis at (6,0).

I hope this answer can help you.

I'm available if you'd like a follow-up.

Regards.
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by GMATGuruNY » Sat Jan 27, 2018 8:48 am
VJesus12 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)
For any quadratic of the form y = x² + bx + c, the SUM of the x-intercepts = -b.
In the quadratic above, b=-5, implying that the sum of the x-intercepts = -(-5) = 5.
Since (-1, 0) is on the curve, one of the x-intercepts is -1.
Since the sum of the x-intercepts must be equal to 5, we get:
-1 + x = 5
x = 6.
Thus, the other x-intercept = (6, 0).

The correct answer is D.
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by Scott@TargetTestPrep » Sun Jul 28, 2019 8:39 am
VJesus12 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)

The OA is the option D
The curve y = x^2 - 5x + t is a parabola. Since it intersects the x-axis at (-1, 0), it means x = -1 is a solution to the equation x^2 - 5x + t = 0. Therefore, we have:

(-1)^2 - 5(-1) + t = 0

1 + 5 + t = 0

t = -6

Since t = -6, we can find the other solution by solving the equation:

x^2 - 5x - 6 = 0

(x + 1)(x - 6) = 0

x = -1 or x = 6

Since we know x = -1 is already a solution, x = 6 is the other solution, and thus the parabola also intersects the x-axis at (6, 0).

Answer: D

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