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Re: If the graph y = 2x2 - 2 intersects line k at (t, 4) and (p, 0), what [#permalink]
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Since the graph \(y = 2x^2 - 2\) intersects line k at (t, 4) and (p, 0), the equation \(y = 2x^2 + 2\) should hold true at both points.

\(4=2t^2+2\)

\(0=2p^2+2 \)

\(t=+1 or-1 \)

\(p=+1 or-1 \)

Since the slope of line k is \(\frac{y^2 - y^1}{x^2-x^1} =\frac{ 0 - 4 }{p - t}\) the maximum value can be obtained when p = -1 and t = 1.

Therefore, the slope is \(\frac{-4}{(-1 - 1)} = 2\).
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Re: If the graph y = 2x2 - 2 intersects line k at (t, 4) and (p, 0), what [#permalink]
Carcass wrote:
Since the graph \(y = 2x^2 - 2\) intersects line k at (t, 4) and (p, 0), the equation \(y = 2x^2 + 2\) should hold true at both points.

\(4=2t^2+2\)

\(0=2p^2+2 \)

\(t=+1 or-1 \)

\(p=+1 or-1 \)

Since the slope of line k is \(\frac{y^2 - y^1}{x^2-x^1} =\frac{ 0 - 4 }{p - t}\) the maximum value can be obtained when p = -1 and t = 1.

Therefore, the slope is \(\frac{-4}{(-1 - 1)} = 2\).


Carcass
How can the equation \(y = 2x^2 + 2\) hold true at both points?
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Re: If the graph y = 2x2 - 2 intersects line k at (t, 4) and (p, 0), what [#permalink]
Expert Reply
KarunMendiratta wrote:
Carcass wrote:
Since the graph \(y = 2x^2 - 2\) intersects line k at (t, 4) and (p, 0), the equation \(y = 2x^2 + 2\) should hold true at both points.

\(4=2t^2+2\)

\(0=2p^2+2 \)

\(t=+1 or-1 \)

\(p=+1 or-1 \)

Since the slope of line k is \(\frac{y^2 - y^1}{x^2-x^1} =\frac{ 0 - 4 }{p - t}\) the maximum value can be obtained when p = -1 and t = 1.

Therefore, the slope is \(\frac{-4}{(-1 - 1)} = 2\).


Carcass
How can the equation \(y = 2x^2 + 2\) hold true at both points?


That is the official explanation. If you have more insights sri feel free to explain :)
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