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Re: If y = -x2 + 1 intersects line k at (t, 3) and (p, 0), what is the max [#permalink]
1
\(y=−x^2+1\)
to get t, substitute the coordinates into the eq
\(-3 = -t^2+1\)
\(t = +-2\)

to get p, substitute the coordinates into the eq
\(0 = -p^2+1\)
\(p = +-1\)

if you draw out the 4 possible points, you will see the max gradient can be obtained with t = -2 and p = -1
gradient \(= (-3-0)/(-2-(-1))\)
= 3, hence the answer is C
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Re: If y = -x2 + 1 intersects line k at (t, 3) and (p, 0), what is the max [#permalink]
KarunMendiratta wrote:
Carcass wrote:
If \(y = -x^2 + 1\) intersects line k at (t, 3) and (p, 0), what is the maximum possible slope for line k?

(A) -3
(B) 2
(C) 3
(D) 4
(E) 9


Graph of \(y = -x^2 + 1\) will never have y = 3 as its maximum (vertex) lies at (0,1)
Carcass Can you please check?


Carcass

The stem has been edited I guess!
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If y = -x2 + 1 intersects line k at (t, 3) and (p, 0), what is the max [#permalink]
Expert Reply
KarunMendiratta wrote:
KarunMendiratta wrote:
Carcass wrote:
If \(y = -x^2 + 1\) intersects line k at (t, 3) and (p, 0), what is the maximum possible slope for line k?

(A) -3
(B) 2
(C) 3
(D) 4
(E) 9


Graph of \(y = -x^2 + 1\) will never have y = 3 as its maximum (vertex) lies at (0,1)
Carcass Can you please check?


Carcass

The stem has been edited I guess!


Fixed sir. Thank you and sorry for the inconvenience.

I edited on my own. I guess the book was wrong

take a look now , please
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