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Point \(P\) lies on the equation \(y=x^2−1\) and Point \(Q

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by AAPL » Wed Jun 26, 2019 6:23 am
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Point \(P\) lies on the equation \(y=x^2−1\) and Point \(Q\) lies on the equation \(y=−x^2+1\). Find the difference between the minimum y coordinate value of Point \(P\) and the maximum \(y\) coordinate value of Point \(Q\).

A. -2
B. -1
C. 0
D. 1
E. 2

OA A
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by Scott@TargetTestPrep » Mon Jul 01, 2019 4:09 pm
AAPL wrote:Princeton Review

Point \(P\) lies on the equation \(y=x^2−1\) and Point \(Q\) lies on the equation \(y=−x^2+1\). Find the difference between the minimum y coordinate value of Point \(P\) and the maximum \(y\) coordinate value of Point \(Q\).

A. -2
B. -1
C. 0
D. 1
E. 2

OA A
We see that y = x^2 - 1 is a parabola opening upward, and y = -x^2 + 1 is a parabola opening downward. Therefore, the lowest point of y = x^2 - 1 is its vertex, which is (0, -1). So the minimum y-coordinate of point P is -1. Similarly, the highest point of y = -x^2 + 1 is its vertex, which (0, 1). So the maximum y-coordinate of point Q is 1. Therefore, the difference is (-1) - 1 = -2.

Answer: A

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by deloitte247 » Fri Jul 05, 2019 8:32 am
$$y=x^2-1$$
$$y+1=x^2\ \ \ \ \ \ \ \ \ \ \left(getting\ the\ intercept\ by\ putting\ x=0\right)$$
$$y+1=0\ \ \ \ \ \ \ \ \ $$
$$y=-1.\ \ Therefore,\ \min\ \left(y\right)\ =\ -1$$

$$y=x^2+1$$
$$y-1=x^2$$
$$y-1=0$$
$$y=1.\ \ \ Therefore,\ \max\left(y\right)=1$$

$$Difference=\minimum-\maximum$$
$$Difference=-1-1=-2\ \ \ \ \ \ \ \left(Answer=option\ A\right)$$
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by Ian Stewart » Sat Jul 06, 2019 4:18 am
If y = x^2 - 1, then since x^2 is at least zero, the smallest possible value of y is -1. If y = -x^2 + 1, then since -x^2 is at most zero, the largest possible value of y is 1. So the answer is -1 - 1 = -2.
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