Secondary 3 Elementary Mathematics Algebra Functions Quiz

Secondary 3 Elementary Mathematics Quiz - Algebra Functions

Answer Key

Total Marks: 40

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Section A: Quadratic Functions and Transformations

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1. Express $y = x^2 + 6x + 11$ in the form $y = (x-p)^2 + q$. Hence state the coordinates of the minimum point. [2 marks]

Method – Completing the square:

This technique rewrites a quadratic in vertex form $y = (x-p)^2 + q$, where $(p, q)$ gives the turning point. We use the identity $(x+p)^2 = x^2 + 2px + p^2$.

• Take the coefficient of $x$, which is $6$, halve it to get $3$, then square it: $3^2 = 9$.

$y = x^2 + 6x + 9 + 11 - 9$

$y = (x+3)^2 + 2$

So $p = -3$ and $q = 2$.

Explanation: The term $(x+3)^2$ is always $\geq 0$, with minimum value $0$ when $x = -3$. Thus the minimum value of $y$ is $0 + 2 = 2$.

Answer: $y = (x+3)^2 + 2$; Minimum point: $(-3, 2)$ [1 mark for correct completed-square form; 1 mark for correct coordinates]

Common mistake: Writing $(x-3)^2$ instead of $(x+3)^2$; forgetting to subtract the $9$ to keep the expression equivalent.

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2. The graph of $y = -(x-3)^2 + 5$ has a maximum point at $A$.

(a) Coordinates of $A$ [1 mark]

Method: For $y = -(x-p)^2 + q$, the negative sign means the parabola opens downwards, so $(p, q)$ is a maximum point.

Comparing with $y = -(x-3)^2 + 5$: we have $p = 3$, $q = 5$.

Answer: $A$ $(3, 5)$

(b) Equation of the line of symmetry [1 mark]

Method: The line of symmetry passes through the turning point $x = p$.

Answer: $x = 3$

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3. Sketch the graph of $y = (x-1)(x-5)$, showing clearly the coordinates of the $x$-intercepts and the $y$-intercept. [3 marks]

Method: This is the factorised form $y = (x-a)(x-b)$ where roots are $x = a$ and $x = b$.

• $x$-intercepts: Set $y = 0$: $(x-1)(x-5) = 0$, so $x = 1$ or $x = 5$.

  • Points: $(1, 0)$ and $(5, 0)$ [1 mark]

• $y$-intercept: Set $x = 0$: $y = (0-1)(0-5) = (-1)(-5) = 5$.

  • Point: $(0, 5)$ [1 mark]

• Shape: Since the coefficient of $x^2$ (when expanded: $y = x^2 - 6x + 5$) is positive, the parabola opens upwards with a minimum point between the roots. [1 mark for correct shape with minimum]

Mark breakdown:

• Correct shape (U-shaped parabola): 1 mark
• Both $x$-intercepts correctly labelled: 1 mark
• $y$-intercept correctly labelled: 1 mark

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4. The graph of $y = x^2 + bx + c$ passes through $(0, 8)$ and $(4, 0)$. Find $b$ and $c$. [2 marks]

Method: Substitute the points into the equation to form simultaneous equations.

From $(0, 8)$: When $x = 0, y = 8$: $8 = 0^2 + b(0) + c$ $c = 8$ [1 mark]

From $(4, 0)$: When $x = 4, y = 0$: $0 = 16 + 4b + 8$ $0 = 24 + 4b$ $4b = -24$ $b = -6$ [1 mark]

Answer: $b = -6$, $c = 8$

Verification: Equation is $y = x^2 - 6x + 8 = (x-2)(x-4)$. Check: $y=0$ when $x=4$ ✓ and $y=8$ when $x=0$ ✓.

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5. Describe the transformation that maps $y = x^2$ onto $y = x^2 - 4x + 6$. [1 mark]

Method: Complete the square for the target equation.

$y = x^2 - 4x + 6 = (x-2)^2 + 4 - 4 + 6 = (x-2)^2 + 2$

Wait – let me redo: $(x-2)^2 = x^2 - 4x + 4$, so: $y = (x-2)^2 + 6 - 4 = (x-2)^2 + 2$

So $y = (x-2)^2 + 2$.

Transformation: From $y = x^2$ to $y = (x-2)^2 + 2$:

• Replace $x$ with $(x-2)$: shift 2 units to the right (horizontal translation)
• Add $2$: shift 2 units up (vertical translation)

Answer: Translation by vector $\begin{pmatrix} 2 \\ 2 \end{pmatrix}$ (or "2 units to the right and 2 units up")

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Section B: Power Functions and Their Graphs

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6. Sketch the graph of $y = x^{-1} = \frac{1}{x}$ for $x \neq 0$. [2 marks]

Key properties:

• For $x > 0$: as $x \to 0^+$, $y \to +\infty$; as $x \to +\infty$, $y \to 0^+$
• For $x < 0$: as $x \to 0^-$, $y \to -\infty$; as $x \to -\infty$, $y \to 0^-$
• No $x$- or $y$-intercept (curve never touches axes)

Asymptotes: [1 mark]

• $x = 0$ (the $y$-axis) is a vertical asymptote
• $y = 0$ (the $x$-axis) is a horizontal asymptote

Shape: [1 mark]

• Two separate branches in quadrants 1 and 3, symmetric about origin (rotational symmetry of order 2)
• Point $(1, 1)$ and $(-1, -1)$ lie on curve

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7. Curve $y = x^2$ undergoes stretch scale factor 3 parallel to $y$-axis, then translation $\begin{pmatrix} 0 \\ -2 \end{pmatrix}$. [2 marks]

Method:

• Stretch SF 3 parallel to $y$-axis: Replace $y$ with $\frac{y}{3}$, or equivalently multiply RHS by 3.

  • New equation: $y = 3x^2$

• Translation by $\begin{pmatrix} 0 \\ -2 \end{pmatrix}$: Replace $y$ with $(y + 2)$, or subtract 2 from RHS.

  • New equation: $y = 3x^2 - 2$

Answer: $y = 3x^2 - 2$ [1 mark for each transformation correctly applied; deduct 1 if order wrong or arithmetic error]

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8. Sketch $y = x^2$ and $y = x^3$ on $-2 \leq x \leq 2$. [2 marks]

$x$	$x^2$	$x^3$
$-2$	$4$	$-8$
$-1$	$1$	$-1$
$0$	$0$	$0$
$1$	$1$	$1$
$2$	$4$	$8$

$y = x^2$ (parabola): [1 mark]

• U-shaped, minimum at origin, symmetric about $y$-axis
• Passes through $(-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4)$

$y = x^3$ (cubic): [1 mark]

• Passes through origin, always increasing
• For $x < 0$: negative values, flatter near origin, steeper for large $|x|$
• Passes through $(-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8)$
• Point of inflection at origin (gradient changes from decreasing to increasing)

Important: Label which curve is which. Curves intersect at $x = 0$ and $x = 1$ (since $x^2 = x^3 \Rightarrow x^2(1-x) = 0$).

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9. Find $x$ such that $x^{-2} \geq \frac{1}{4}$. [1 mark]

Method: $x^{-2} = \frac{1}{x^2}$, so: $\frac{1}{x^2} \geq \frac{1}{4}$

Since both sides are positive (and $x^2 > 0$ for $x \neq 0$), we can cross-multiply safely: $4 \geq x^2$ or equivalently $x^2 \leq 4$

This gives $-2 \leq x \leq 2$, but we must exclude $x = 0$ where the original expression is undefined.

Answer: $-2 \leq x < 0$ or $0 < x \leq 2$ (or $-2 \leq x \leq 2, x \neq 0$)

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10. Find $k$ where $y = kx^2$ passes through $(3, 18)$. [1 mark]

Method: Substitute the point into the equation. $18 = k \times 3^2 = 9k$ $k = 2$

Answer: $k = 2$

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Section C: Exponential Functions

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11. Sketch $y = 2^x$ for $-3 \leq x \leq 3$. [2 marks]

Key values:

$x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$y$	$\frac{1}{8}$	$\frac{1}{4}$	$\frac{1}{2}$	$1$	$2$	$4$	$8$

Properties: [1 mark for shape + key points]

• Always positive: $y > 0$ for all $x$
• Strictly increasing function
• $y$-intercept at $(0, 1)$ [1 mark]
• As $x \to -\infty$, $y \to 0$ (negative $x$-axis is horizontal asymptote)
• Rapid growth as $x$ increases

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12. Graph of $y = 3^x$ reflected in the $x$-axis. [1 mark]

Method: Reflection in $x$-axis: replace $y$ with $-y$, or multiply RHS by $-1$.

Answer: $y = -3^x$

Note: This is not the same as $y = (-3)^x$ or $y = 3^{-x}$. The entire graph is flipped vertically.

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13. Solve $5 \times 2^x = 80$. [2 marks]

Method: $2^x = \frac{80}{5} = 16$ [1 mark] $2^x = 2^4$ $x = 4$ [1 mark]

Answer: $x = 4$

Common mistake: Trying to "take logs" without simplifying first, or writing $x = \sqrt[5]{16}$.

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14. Bacteria population $P = 100 \times 2^t$.

(a) Population at $t = 0$: [1 mark] $P = 100 \times 2^0 = 100 \times 1 = 100$

Answer: 100 bacteria

(b) Time to reach 6400: [2 marks] $100 \times 2^t = 6400$ $2^t = 64$ [1 mark] $2^t = 2^6$ $t = 6$ [1 mark]

Answer: 6 hours

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15. Find $a$ where $y = a^x$ passes through $(0, 1)$ and $(2, 25)$. [2 marks]

Check $(0, 1)$: $a^0 = 1$ ✓ (true for any $a > 0, a \neq 1$; confirms form is valid)

From $(2, 25)$: $a^2 = 25$ [1 mark] $a = 5$ (we take positive root since $a > 1$ for standard exponential growth) [1 mark]

Reject $a = -5$ since exponential base must be positive.

Answer: $a = 5$

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Section D: Gradients and Curves; Composite Functions

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16. Point $P(2, 5)$ on curve $y = x^2 + 1$.

(a) Gradient of chord $PQ$ where $Q = (2.1, 2.1^2 + 1) = (2.1, 5.41)$. [2 marks]

Method: $\text{Gradient} = \frac{y_Q - y_P}{x_Q - x_P} = \frac{5.41 - 5}{2.1 - 2} = \frac{0.41}{0.1} = 4.1$ [1 mark for substitution; 1 mark for answer]

Answer: 4.1

(b) Estimate gradient of tangent at $P$: [1 mark]

Method: As $Q$ gets closer to $P$ (i.e., as $x_Q \to 2$), the chord gradient approaches the tangent gradient. We could use a smaller interval, e.g., $Q = (2.01, 5.0401)$, giving gradient $\frac{0.0401}{0.01} = 4.01$, which approaches 4.

Answer: The gradient of the tangent is estimated by taking $Q$ closer and closer to $P$ (limiting process) — the tangent gradient is approximately 4 (exact: limit as $h \to 0$ of $\frac{(2+h)^2+1-5}{h} = \frac{4h+h^2}{h} = 4+h \to 4$).

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17. Estimate gradient of $y = x^2 - 3x + 5$ at $x = 4$ using tangents. [3 marks]

Expected graph features (from image placeholder):

• Curve is parabola $y = x^2 - 3x + 5$
• At $x = 4$: $y = 16 - 12 + 5 = 9$, so point is $(4, 9)$
• Tangent line drawn on graph

Method using graph:

1. Identify two points on the tangent line from the drawn tangent. Expected: the tangent might pass through approximately $(3, 2)$ and $(5, 16)$ or similar points with clear integer coordinates. [1 mark for identifying points on tangent]

2. Calculate gradient: Using expected visible points on tangent: if tangent passes through $(3, 2)$ and $(5, 16)$: $\text{Gradient} = \frac{16 - 2}{5 - 3} = \frac{14}{2} = 7$?

   Wait — let me verify with calculus for the expected answer key. The exact derivative is $\frac{dy}{dx} = 2x - 3$, so at $x = 4$: gradient = $5$.

   So the tangent should have gradient 5. If passing through $(4, 9)$, points with gradient 5: $(2, -1)$ and $(6, 19)$ — but these may be off the visible scale.

   From image: likely gradient triangle shows "rise 5, run 1" or "rise 10, run 2".

Expected student approach: Draw/acquire two points on tangent, calculate $\frac{\Delta y}{\Delta x}$. [1 mark]

Verification with expected values: If image shows gradient triangle with rise ≈ 5 and run ≈ 1, or using points like $(3.5, 6.5)$ and $(4.5, 11.5)$: gradient = $\frac{5}{1} = 5$. [1 mark for answer in range 4.5–5.5]

Answer: 5 (acceptable range: 4.5 to 5.5 due to estimation)

Exact verification: $\frac{dy}{dx} = 2x - 3 = 8 - 3 = 5$ when $x = 4$.

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18. $f(x) = 2x + 3$ and $g(x) = x^2 - 1$.

(a) Find $f(g(2))$. [2 marks]

Method - work from inside out:

• First find $g(2)$: $g(2) = 2^2 - 1 = 4 - 1 = 3$ [1 mark]

• Then $f(g(2)) = f(3) = 2(3) + 3 = 6 + 3 = 9$ [1 mark]

Answer: 9

(b) Find $g(f(x))$. [2 marks]

Method: $g(f(x)) = g(2x+3) = (2x+3)^2 - 1$ [1 mark for substitution] $= 4x^2 + 12x + 9 - 1$ $= 4x^2 + 12x + 8$ [1 mark for expansion]

Or factored: $= 4(x^2 + 3x + 2) = 4(x+1)(x+2)$

Answer: $4x^2 + 12x + 8$

Common mistake: Computing $f(g(x))$ instead (which would be $2(x^2-1)+3 = 2x^2+1$). Note that $f \circ g \neq g \circ f$ in general.

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19. $h(x) = \frac{3}{x-2}$ for $x \neq 2$. Find $h^{-1}(1)$. [2 marks]

Method - two approaches:

Approach 1: Find inverse function first. Let $y = \frac{3}{x-2}$. Swap $x$ and $y$: $x = \frac{3}{y-2}$ $x(y-2) = 3$ $y - 2 = \frac{3}{x}$ $y = \frac{3}{x} + 2 = \frac{3+2x}{x}$

So $h^{-1}(x) = \frac{3}{x} + 2$.

Then $h^{-1}(1) = 3 + 2 = 5$. [1 mark for finding inverse; 1 mark for substitution]

Approach 2: Use definition of inverse. $h^{-1}(1) = a$ means $h(a) = 1$.

So $\frac{3}{a-2} = 1$ $3 = a - 2$ $a = 5$ [2 marks if correct, or 1 mark for setting up equation]

Answer: $h^{-1}(1) = 5$

Verification: $h(5) = \frac{3}{5-2} = \frac{3}{3} = 1$ ✓

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20. Range of $k$ for which $y = 2x + k$ intersects $y = x^2 - 4x + 7$ at two distinct points. [3 marks]

Method: For intersection, set equations equal: $2x + k = x^2 - 4x + 7$ $0 = x^2 - 6x + (7 - k)$ [1 mark for rearranging to standard form]

For two distinct real solutions, discriminant $b^2 - 4ac > 0$.

Here $a = 1$, $b = -6$, $c = 7-k$: $(-6)^2 - 4(1)(7-k) > 0$ $36 - 28 + 4k > 0$ [1 mark for discriminant condition] $8 + 4k > 0$ $4k > -8$ $k > -2$ [1 mark]

Answer: $k > -2$

Geometric interpretation: The line $y = 2x + k$ has fixed gradient 2 but varying $y$-intercept $k$. As $k$ increases, the line shifts up. When $k = -2$, the line is tangent to the parabola (one point of contact). For $k > -2$, it cuts through two points.

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END OF ANSWER KEY