TMUA Functions & graphs questions

Answer: C · Method: Count roots using turning points

1. The curve $y = x^3 - 3x$ has turning points at $x = \pm 1$: a maximum at $(-1, 2)$ and a minimum at $(1, -2)$.
2. A horizontal line $y = k$ meets the curve three times exactly when $-2 < k < 2$, so $k = -1, 0, 1$.

Why the tempting wrong answers are wrong

• E: At $k = \pm 2$ the line touches a turning point, giving only two distinct solutions.

Answer: B · Method: Compose functions, then use the sum of roots

1. $f(g(x)) = 2(x^2 - 1) + 3 = 2x^2 + 1$ and $g(f(x)) = (2x + 3)^2 - 1 = 4x^2 + 12x + 8$.
2. Setting them equal: $2x^2 + 12x + 7 = 0$. Its discriminant $144 - 56 > 0$, so there are two real roots, with sum $-\tfrac{12}{2} = -6$.

Why the tempting wrong answers are wrong

• D: $\tfrac{7}{2}$ is the product of the roots, $\tfrac{c}{a}$, not the sum $-\tfrac{b}{a}$.

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

1. Complete the square in $x$: $y = (x - a)^2 + a^2 - 4a + 1$, so the minimum point is $(a,\; a^2 - 4a + 1)$.
2. Now complete the square in $a$: $a^2 - 4a + 1 = (a - 2)^2 - 3$, which is least when $a = 2$. The answer is $-3$.

Why the tempting wrong answers are wrong

• A: Check the completed square: $(a - 2)^2 = a^2 - 4a + 4$, so you need $-3$ to get back to $+1$.

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

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

Why the tempting wrong answers are wrong

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

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

1. Where they meet, $x^2 + 2x + 7 = 3x + c$, so $x^2 - x + (7 - c) = 0$.
2. A tangent touches the curve once, so this quadratic has a repeated root: its discriminant is $0$.
3. $1 - 4(7 - c) = 0$, so $c = 7 - \tfrac{1}{4} = \dfrac{27}{4}$.

Why the tempting wrong answers are wrong

• A: Subtract the line from the curve: the $x$ coefficient becomes the difference of the gradients.
• D: Check the sign when you rearrange $b^2 - 4ac = 0$ for $c$.

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

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

Why the tempting wrong answers are wrong

• A: Subtract the line from the curve: the $x$ coefficient becomes the difference of the gradients.
• D: Check the sign when you rearrange $b^2 - 4ac = 0$ for $c$.

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

1. Complete the square in $x$: $y = (x - a)^2 + 2a^2 + 8a$, so the minimum point is at height $2a^2 + 8a$.
2. Minimise that over $a$: it's a U-shaped quadratic in $a$, least at $a = -2$.
3. The least value is $0 - \dfrac{64}{8} = -8$.

Why the tempting wrong answers are wrong

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

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

1. Complete the square in $x$: $y = (x - a)^2 + a^2 + 8a - 1$, so the minimum point is at height $a^2 + 8a - 1$.
2. Minimise that over $a$: it's a U-shaped quadratic in $a$, least at $a = -4$.
3. The least value is $-1 - \dfrac{64}{4} = -17$.

Why the tempting wrong answers are wrong

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

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

1. Complete the square in $x$: $y = (x - a)^2 + a^2 + 2a + 6$, so the minimum point is at height $a^2 + 2a + 6$.
2. Minimise that over $a$: it's a U-shaped quadratic in $a$, least at $a = -1$.
3. The least value is $6 - \dfrac{4}{4} = 5$.

Why the tempting wrong answers are wrong

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

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

1. Complete the square in $x$: $y = (x - a)^2 + 2a^2 - 6a - 4$, so the minimum point is at height $2a^2 - 6a - 4$.
2. Minimise that over $a$: it's a U-shaped quadratic in $a$, least at $a = \dfrac{3}{2}$.
3. The least value is $-4 - \dfrac{36}{8} = -\dfrac{17}{2}$.

Why the tempting wrong answers are wrong

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

Answer: C · Method: Track a point through graph transformations

1. Replacing $x$ with $x - 4$ moves the graph right by $4$: the turning point moves to $x = 3$.
2. Multiplying by $2$ stretches vertically: the height becomes $2 \times 5 = 10$.
3. Then subtract $3$, giving $(3, 7)$.

Why the tempting wrong answers are wrong

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

Answer: B · Method: Track a point through graph transformations

1. Replacing $x$ with $x - 3$ moves the graph right by $3$: the turning point moves to $x = 7$.
2. Multiplying by $2$ stretches vertically: the height becomes $2 \times 1 = 2$.
3. Then add $4$, giving $(7, 6)$.

Why the tempting wrong answers are wrong

• A: The stretch applies first, then the vertical shift.
• C: Don't forget the vertical stretch by $2$.
• E: $f(x - 3)$ shifts the graph right by $3$.