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9709 P12 - Jun 2018 - Q2 - 5 marks
322
The equation of a curve is \(y = x^2 - 6x + k\), where \(k\) is a constant.
(i) Find the set of values of \(k\) for which the whole of the curve lies above the \(x\)-axis.
(ii) Find the value of \(k\) for which the line \(y + 2x = 7\) is a tangent to the curve.
Solution
(i) To find when the curve \(y = x^2 - 6x + k\) lies entirely above the \(x\)-axis, the quadratic must have no real roots. This occurs when the discriminant \(b^2 - 4ac < 0\). Here, \(a = 1\), \(b = -6\), and \(c = k\). The discriminant is \((-6)^2 - 4(1)(k) = 36 - 4k\). For the curve to lie above the \(x\)-axis, \(36 - 4k < 0\), which simplifies to \(k > 9\).
(ii) For the line \(y + 2x = 7\) to be tangent to the curve, it must touch the curve at exactly one point. Substitute \(y = 7 - 2x\) into the curve equation: \(7 - 2x = x^2 - 6x + k\). Rearrange to form a quadratic: \(x^2 - 4x + (k - 7) = 0\). For tangency, the discriminant must be zero: \((-4)^2 - 4(1)(k - 7) = 0\). Simplifying gives \(16 - 4k + 28 = 0\), leading to \(44 = 4k\), so \(k = 11\).
