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ESAT Maths 2 Functions & graphs questions

12 original ESAT Maths 2 functions & graphs questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Composite and inverse functions: work from the inside outTangent means a repeated root (discriminant = 0)Complete the square, then minimise againTrack a point through graph transformations

Question 1ESAT Maths 2Warm-up

The functions ff and gg are defined by f(x)=−3x−5f(x) = -3x - 5 and g(x)=x2+2g(x) = x^2 + 2. What is gf(3)gf(3)?
  1. A−154-154
  2. B−38-38
  3. C197197
  4. D198198
  5. E199199
  6. F200200
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Answer: D · Method: Composite and inverse functions: work from the inside out

  1. f(3)=−14f(3) = -14, then g(−14)=198g(-14) = 198.

Why the tempting wrong answers are wrong

  • B: gf(x)gf(x) means apply ff first, then gg.

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Question 2ESAT Maths 2Warm-up

The functions ff and gg are defined by f(x)=3x+4f(x) = 3x + 4 and g(x)=x2+4g(x) = x^2 + 4. What is f−1(g(2))f^{-1}(g(2))?
  1. A128\dfrac{1}{28}
  2. B13\dfrac{1}{3}
  3. C43\dfrac{4}{3}
  4. D73\dfrac{7}{3}
  5. E103\dfrac{10}{3}
  6. F2828
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Answer: C · Method: Composite and inverse functions: work from the inside out

  1. g(2)=8g(2) = 8. The inverse undoes ff: f−1(x)=x−43f^{-1}(x) = \dfrac{x - 4}{3}.
  2. So f−1(8)=43f^{-1}(8) = \dfrac{4}{3}.

Why the tempting wrong answers are wrong

  • A: f−1f^{-1} is the inverse function, not 1/f1/f.

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Question 3ESAT Maths 2Warm-up

The functions ff and gg are defined by f(x)=4x−3f(x) = 4x - 3 and g(x)=x2+4g(x) = x^2 + 4. What is f−1(g(4))f^{-1}(g(4))?
  1. A177\dfrac{1}{77}
  2. B154\dfrac{15}{4}
  3. C194\dfrac{19}{4}
  4. D234\dfrac{23}{4}
  5. E274\dfrac{27}{4}
  6. F314\dfrac{31}{4}
  7. G7777
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Answer: D · Method: Composite and inverse functions: work from the inside out

  1. g(4)=20g(4) = 20. The inverse undoes ff: f−1(x)=x+34f^{-1}(x) = \dfrac{x + 3}{4}.
  2. So f−1(20)=234f^{-1}(20) = \dfrac{23}{4}.

Why the tempting wrong answers are wrong

  • A: f−1f^{-1} is the inverse function, not 1/f1/f.

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Question 4ESAT Maths 2Standard

The line y=2x+cy = 2x + c is a tangent to the curve y=x2−x+5y = x^2 - x + 5. What is the value of cc?
  1. A12\dfrac{1}{2}
  2. B114\dfrac{11}{4}
  3. C154\dfrac{15}{4}
  4. D194\dfrac{19}{4}
  5. E294\dfrac{29}{4}
Show the answer and worked solution

Answer: B · Method: Tangent means a repeated root (discriminant = 0)

  1. Where they meet, x2−x+5=2x+cx^2 - x + 5 = 2x + c, so x2−3x+(5−c)=0x^2 - 3x + (5 - c) = 0.
  2. A tangent touches the curve once, so this quadratic has a repeated root: its discriminant is 00.
  3. 9−4(5−c)=09 - 4(5 - c) = 0, so c=5−94=114c = 5 - \tfrac{9}{4} = \dfrac{11}{4}.

Why the tempting wrong answers are wrong

  • D: Subtract the line from the curve: the xx coefficient becomes the difference of the gradients.
  • E: Check the sign when you rearrange b2−4ac=0b^2 - 4ac = 0 for cc.

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Question 5ESAT Maths 2Standard

The curve y=x2−2ax+2a2−6a−3y = x^2 - 2ax + 2a^2 - 6a - 3 has a minimum point whose position depends on the constant aa. As aa varies, what is the least possible yy-coordinate of this minimum point?
  1. A−21-21
  2. B−14-14
  3. C−13-13
  4. D−12-12
  5. E−11-11
  6. F−10-10
  7. G−3-3
  8. H66
Show the answer and worked solution

Answer: D · Method: Complete the square, then minimise again

  1. Complete the square in xx: y=(x−a)2+a2−6a−3y = (x - a)^2 + a^2 - 6a - 3, so the minimum point is at height a2−6a−3a^2 - 6a - 3.
  2. Minimise that over aa: it's a U-shaped quadratic in aa, least at a=3a = 3.
  3. The least value is −3−364=−12-3 - \dfrac{36}{4} = -12.

Why the tempting wrong answers are wrong

  • G: That's the value when a=0a = 0, but aa can vary to make it smaller.
  • H: Check the sign of the constant when you complete the square in aa.

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Question 6ESAT Maths 2Standard

The curve y=f(x)y = f(x) has a maximum point at (−5,5)(-5, 5) and no other stationary points. What are the coordinates of the maximum point of y=3f(x−1)−1y = 3f(x - 1) - 1?
  1. A(−6,14)(-6, 14)
  2. B(−4,−16)(-4, -16)
  3. C(−4,12)(-4, 12)
  4. D(−4,4)(-4, 4)
  5. E(−4,14)(-4, 14)
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Answer: E · Method: Track a point through graph transformations

  1. Replacing xx with x−1x - 1 moves the graph right by 11: the turning point moves to x=−4x = -4.
  2. Multiplying by 33 stretches vertically: the height becomes 3×5=153 \times 5 = 15.
  3. Then subtract 11, giving (−4,14)(-4, 14).

Why the tempting wrong answers are wrong

  • A: f(x−1)f(x - 1) shifts the graph right by 11.
  • C: The stretch applies first, then the vertical shift.
  • D: Don't forget the vertical stretch by 33.

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Question 7ESAT Maths 2Standard

The functions ff and gg are defined by f(x)=3x+2f(x) = 3x + 2 and g(x)=x2+1g(x) = x^2 + 1. What is the positive solution of fg(x)=32fg(x) = 32?
  1. A11
  2. B22
  3. C33
  4. D44
  5. E55
  6. F66
  7. G99
  8. H1010
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Answer: C · Method: Composite and inverse functions: work from the inside out

  1. fg(x)=3(x2+1)+2=32fg(x) = 3(x^2 + 1) + 2 = 32.
  2. So x2+1=10x^2 + 1 = 10, x2=9x^2 = 9, and x=3x = 3.

Why the tempting wrong answers are wrong

  • G: That's x2x^2. Take the square root.

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Question 8ESAT Maths 2Standard

The line y=−2x+cy = -2x + c is a tangent to the curve y=x2+6x+2y = x^2 + 6x + 2. What is the value of cc?
  1. A−30-30
  2. B−15-15
  3. C−14-14
  4. D−13-13
  5. E−2-2
  6. F1818
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Answer: C · Method: Tangent means a repeated root (discriminant = 0)

  1. Where they meet, x2+6x+2=−2x+cx^2 + 6x + 2 = -2x + c, so x2+8x+(2−c)=0x^2 + 8x + (2 - c) = 0.
  2. A tangent touches the curve once, so this quadratic has a repeated root: its discriminant is 00.
  3. 64−4(2−c)=064 - 4(2 - c) = 0, so c=2−644=−14c = 2 - \tfrac{64}{4} = -14.

Why the tempting wrong answers are wrong

  • E: Subtract the line from the curve: the xx coefficient becomes the difference of the gradients.
  • F: Check the sign when you rearrange b2−4ac=0b^2 - 4ac = 0 for cc.

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Question 9ESAT Maths 2Standard

The line y=−x+cy = -x + c is a tangent to the curve y=x2+6y = x^2 + 6. What is the value of cc?
  1. A194\dfrac{19}{4}
  2. B112\dfrac{11}{2}
  3. C234\dfrac{23}{4}
  4. D254\dfrac{25}{4}
  5. E274\dfrac{27}{4}
  6. F314\dfrac{31}{4}
Show the answer and worked solution

Answer: C · Method: Tangent means a repeated root (discriminant = 0)

  1. Where they meet, x2+6=−x+cx^2 + 6 = -x + c, so x2+x+(6−c)=0x^2 + x + (6 - c) = 0.
  2. A tangent touches the curve once, so this quadratic has a repeated root: its discriminant is 00.
  3. 1−4(6−c)=01 - 4(6 - c) = 0, so c=6−14=234c = 6 - \tfrac{1}{4} = \dfrac{23}{4}.

Why the tempting wrong answers are wrong

  • D: Check the sign when you rearrange b2−4ac=0b^2 - 4ac = 0 for cc.

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Question 10ESAT Maths 2Standard

The curve y=x2−2ax+2a2−2a+4y = x^2 - 2ax + 2a^2 - 2a + 4 has a minimum point whose position depends on the constant aa. As aa varies, what is the least possible yy-coordinate of this minimum point?
  1. A−1-1
  2. B11
  3. C22
  4. D33
  5. E44
  6. F55
  7. G66
  8. H77
Show the answer and worked solution

Answer: D · Method: Complete the square, then minimise again

  1. Complete the square in xx: y=(x−a)2+a2−2a+4y = (x - a)^2 + a^2 - 2a + 4, so the minimum point is at height a2−2a+4a^2 - 2a + 4.
  2. Minimise that over aa: it's a U-shaped quadratic in aa, least at a=1a = 1.
  3. The least value is 4−44=34 - \dfrac{4}{4} = 3.

Why the tempting wrong answers are wrong

  • E: That's the value when a=0a = 0, but aa can vary to make it smaller.
  • F: Check the sign of the constant when you complete the square in aa.

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Question 11ESAT Maths 2Standard

The curve y=x2−2ax+3a2+4a+1y = x^2 - 2ax + 3a^2 + 4a + 1 has a minimum point whose position depends on the constant aa. As aa varies, what is the least possible yy-coordinate of this minimum point?
  1. A−5-5
  2. B−3-3
  3. C−2-2
  4. D−1-1
  5. E00
  6. F11
  7. G22
  8. H33
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Answer: D · Method: Complete the square, then minimise again

  1. Complete the square in xx: y=(x−a)2+2a2+4a+1y = (x - a)^2 + 2a^2 + 4a + 1, so the minimum point is at height 2a2+4a+12a^2 + 4a + 1.
  2. Minimise that over aa: it's a U-shaped quadratic in aa, least at a=−1a = -1.
  3. The least value is 1−168=−11 - \dfrac{16}{8} = -1.

Why the tempting wrong answers are wrong

  • F: That's the value when a=0a = 0, but aa can vary to make it smaller.
  • H: Check the sign of the constant when you complete the square in aa.

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Question 12ESAT Maths 2Standard

The curve y=f(x)y = f(x) has a maximum point at (2,−1)(2, -1) and no other stationary points. What are the coordinates of the minimum point of y=−2f(x−4)+4y = -2f(x - 4) + 4?
  1. A(6,2)(6, 2)
  2. B(6,6)(6, 6)
  3. C(−2,6)(-2, 6)
  4. D(6,3)(6, 3)
  5. E(6,−6)(6, -6)
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Answer: B · Method: Track a point through graph transformations

  1. Replacing xx with x−4x - 4 moves the graph right by 44: the turning point moves to x=6x = 6.
  2. Multiplying by −2-2 stretches vertically and reflects in the xx-axis, turning the maximum into a minimum: the height becomes −2×(−1)=2-2 \times (-1) = 2.
  3. Then add 44, giving (6,6)(6, 6).

Why the tempting wrong answers are wrong

  • C: f(x−4)f(x - 4) shifts the graph right by 44.
  • D: Don't forget the vertical stretch by −2-2.
  • E: The stretch applies first, then the vertical shift.

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