The equation of a curve is $y = (m+1)x^2 - 8x + 3m$, where $m$ is a constant. In the case w

Question

The equation of a curve is $y = (m+1)x^2 - 8x + 3m$, where $m$ is a constant. In the case where $m = 1$, show that the line $y = 1 - 4x$ is a tangent to the curve. Find the coordinates of the point of contact.[3] Find the range of values of $m$ for which the curve is above the line $y = 5$.[5]

Accepted Answer

(a) $(1,-3)$ (b) $m>3$

2025 SCGS PRELIMS P1 Q5

2025 SCGS PRELIMS P1 Q5

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