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Quadratics: Question 7

Syllabus 1.1

Structured AS 8 marks

A function is defined by g(x)=−2x2+12x−7g(x) = -2x^2 + 12x - 7 for real xx.

(a) Express g(x)g(x) in the form −2(x−p)2+q-2(x - p)^2 + q, stating the values of the constants pp and qq. [3]

(b) Write down the coordinates of the maximum point of the graph of y=g(x)y = g(x), explaining how you know this point is a maximum rather than a minimum. [2]

(c) Hence, or otherwise, find the exact solutions of g(x)=0g(x) = 0, giving each answer as a single fraction involving a surd. [3]

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Worked solution

Part (a): Completing the square

Factor −2-2 out of only the terms containing xx:

−2x2+12x−7=−2(x2−6x)−7-2x^2 + 12x - 7 = -2(x^2 - 6x) - 7

Complete the square inside the bracket, using x2−6x=(x−3)2−9x^2-6x = (x-3)^2 - 9:

=−2[(x−3)2−9]−7= -2\big[(x-3)^2 - 9\big] - 7

Multiply through by the −2-2, then combine the constants:

=−2(x−3)2+18−7=−2(x−3)2+11= -2(x-3)^2 + 18 - 7 = -2(x-3)^2 + 11

So g(x)=−2(x−3)2+11g(x) = -2(x-3)^2 + 11, giving p=3p = 3, q=11q = 11.

Check by expanding back: −2(x−3)2+11=−2(x2−6x+9)+11=−2x2+12x−18+11=−2x2+12x−7-2(x-3)^2+11 = -2(x^2-6x+9)+11 = -2x^2+12x-18+11 = -2x^2+12x-7, which matches the original g(x)g(x). ✓

Part (b): Maximum point

In the form −2(x−p)2+q-2(x-p)^2+q, the vertex of the parabola is at (p,q)=(3,11)(p, q) = (3, 11).

Since the coefficient of (x−3)2(x-3)^2 is −2<0-2 < 0, the graph is a downward-opening parabola (a "∩\cap" shape), so this vertex is the highest point on the curve, a maximum, not a minimum.

Part (c): Solving g(x)=0g(x) = 0 exactly

Using the completed-square form:

−2(x−3)2+11=0-2(x-3)^2 + 11 = 0

(x−3)2=112(x-3)^2 = \frac{11}{2}

x−3=±112=±112=±222x - 3 = \pm\sqrt{\frac{11}{2}} = \pm\frac{\sqrt{11}}{\sqrt{2}} = \pm\frac{\sqrt{22}}{2}

x=3±222=6±222x = 3 \pm \frac{\sqrt{22}}{2} = \frac{6 \pm \sqrt{22}}{2}

Check using the quadratic formula. With a=−2a=-2, b=12b=12, c=−7c=-7:

discriminant=122−4(−2)(−7)=144−56=88\text{discriminant} = 12^2 - 4(-2)(-7) = 144 - 56 = 88

x=−12±882(−2)=−12±222−4=12∓2224=6∓222x = \frac{-12 \pm \sqrt{88}}{2(-2)} = \frac{-12 \pm 2\sqrt{22}}{-4} = \frac{12 \mp 2\sqrt{22}}{4} = \frac{6 \mp \sqrt{22}}{2}

This is the same pair of values as 6±222\dfrac{6 \pm \sqrt{22}}{2} (the ±\pm symbol covers both orderings), so the two methods agree.

Numerical check: 22≈4.69\sqrt{22}\approx4.69, so x≈5.35x\approx5.35 or x≈0.65x\approx0.65. Substituting x=5.35x=5.35: g(5.35)=−2(5.35)2+12(5.35)−7≈−57.2+64.2−7≈0g(5.35)=-2(5.35)^2+12(5.35)-7\approx-57.2+64.2-7\approx0. ✓

Final answers

  • (a) g(x)=−2(x−3)2+11g(x) = -2(x-3)^2 + 11, with p=3p=3, q=11q=11
  • (b) Maximum point (3,11)\boxed{(3, 11)}
  • (c) x=6±222x = \boxed{\dfrac{6 \pm \sqrt{22}}{2}}