A Level H1 Mathematics Graphs Coordinate Geometry Quiz

A-Level Maths H1 Quiz - Graphs Coordinate Geometry: Answer Key

Total Marks: 50

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Section A: Equations and Inequalities (Questions 1–5)

Question 1 [4 marks]

Answer: $k < -1$ or $k > 4$

Working:

For the quadratic equation $kx^2 + 4x + k - 3 = 0$ to have two distinct real roots, the discriminant must be positive: $b^2 - 4ac > 0$.

Here, $a = k$, $b = 4$, $c = k - 3$.

$b^2 - 4ac = 4^2 - 4(k)(k - 3) = 16 - 4k^2 + 12k$

For two distinct real roots: $16 - 4k^2 + 12k > 0$ $-4k^2 + 12k + 16 > 0$ $k^2 - 3k - 4 < 0$ $(k - 4)(k + 1) < 0$

The critical values are $k = -1$ and $k = 4$. Testing intervals:

• $k < -1$: $(k-4)(k+1) > 0$ (product of two negatives is positive, but we need negative) — No
• $-1 < k < 4$: $(k-4)(k+1) < 0$ — Yes
• $k > 4$: $(k-4)(k+1) > 0$ — No

Wait, let me recheck. We need $(k - 4)(k + 1) < 0$.

• $k < -1$: both $(k-4)$ and $(k+1)$ are negative, product is positive. Not a solution.
• $-1 < k < 4$: $(k-4)$ is negative, $(k+1)$ is positive, product is negative. Solution.
• $k > 4$: both are positive, product is positive. Not a solution.

So the solution is $-1 < k < 4$.

Correction: The answer is $-1 < k < 4$.

Marking Notes:

• M1: Correctly identifies the discriminant condition ($b^2 - 4ac > 0$)
• M1: Correctly substitutes $a = k$, $b = 4$, $c = k - 3$ into the discriminant
• M1: Correctly simplifies to a quadratic inequality
• A1: Correct final range $-1 < k < 4$

Common Mistake: Forgetting that $k \ne 0$ (if $k = 0$, the equation is linear, not quadratic). However, $k = 0$ is not in the solution range anyway, so this does not affect the answer here.

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Question 2 [4 marks]

Answer: $-3 < x \le 4$

Working:

$\frac{2x - 1}{x + 3} \le 1$

Step 1: Bring all terms to one side. $\frac{2x - 1}{x + 3} - 1 \le 0$

Step 2: Combine into a single fraction. $\frac{2x - 1 - (x + 3)}{x + 3} \le 0$ $\frac{2x - 1 - x - 3}{x + 3} \le 0$ $\frac{x - 4}{x + 3} \le 0$

Step 3: The critical values are $x = 4$ (where numerator is zero) and $x = -3$ (where denominator is zero, undefined).

Step 4: Test intervals:

• $x < -3$: numerator negative, denominator negative, fraction positive. Not a solution.
• $-3 < x < 4$: numerator negative, denominator positive, fraction negative. Solution.
• $x > 4$: numerator positive, denominator positive, fraction positive. Not a solution.

Step 5: Include $x = 4$ (since $\le 0$ allows equality), exclude $x = -3$ (undefined).

Answer: $-3 < x \le 4$

Marking Notes:

• M1: Subtracts 1 from both sides to form a single fraction
• M1: Correctly simplifies the numerator
• M1: Identifies critical values and tests intervals correctly
• A1: Correct final answer with correct inequality signs

Common Mistake: Multiplying both sides by $(x+3)$ without considering its sign. Since $(x+3)$ can be negative, this reverses the inequality. Always bring everything to one side first.

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Question 3 [4 marks]

Answer: $0 \le p < 5$

Working:

For $px^2 + 2px + 5$ to be always positive for all real $x$:

Case 1: $p = 0$

If $p = 0$, the expression becomes $5$, which is always positive. So $p = 0$ is a valid solution.

Case 2: $p \ne 0$

For a quadratic to be always positive, we need:

1. $p > 0$ (the parabola opens upwards)
2. Discriminant $< 0$ (no real roots, so it never touches the x-axis)

Discriminant: $b^2 - 4ac = (2p)^2 - 4(p)(5) = 4p^2 - 20p = 4p(p - 5)$

For discriminant $< 0$: $4p(p - 5) < 0$ $0 < p < 5$

Combining with $p > 0$: $0 < p < 5$

Combining both cases: $0 \le p < 5$

Marking Notes:

• M1: Considers the case $p = 0$ separately
• M1: States conditions for always positive quadratic ($p > 0$ and discriminant $< 0$)
• M1: Correctly computes and solves the discriminant inequality
• A1: Correct final answer $0 \le p < 5$

Common Mistake: Forgetting to check $p = 0$ as a special case. Many students only consider the quadratic case.

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Question 4 [4 marks]

Answer: $x = 2, y = 0$ and $x = 3, y = 2$

Working:

Substitute $y = 2x - 4$ into $y = x^2 - 3x + 2$:

$2x - 4 = x^2 - 3x + 2$ $x^2 - 3x + 2 - 2x + 4 = 0$ $x^2 - 5x + 6 = 0$ $(x - 2)(x - 3) = 0$

So $x = 2$ or $x = 3$.

When $x = 2$: $y = 2(2) - 4 = 0$ When $x = 3$: $y = 2(3) - 4 = 2$

Answer: $(2, 0)$ and $(3, 2)$

Marking Notes:

• M1: Substitutes the linear equation into the quadratic
• M1: Rearranges to form a quadratic equation in $x$
• M1: Factorises or solves the quadratic correctly
• A1: Both pairs of solutions correct

Common Mistake: Substituting the wrong way round, or making arithmetic errors when rearranging. Always double-check by substituting both solutions back into both original equations.

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Question 5 [3 marks]

Answer: $10 < x < 30$

Working:

We need $P(x) > 0$: $-2x^2 + 80x - 600 > 0$

Divide by $-2$ (reversing the inequality): $x^2 - 40x + 300 < 0$ $(x - 10)(x - 30) < 0$

Critical values: $x = 10$ and $x = 30$.

Testing intervals:

• $x < 10$: both factors negative, product positive. Not a solution.
• $10 < x < 30$: $(x-10)$ positive, $(x-30)$ negative, product negative. Solution.
• $x > 30$: both factors positive, product positive. Not a solution.

Answer: $10 < x < 30$

Marking Notes:

• M1: Sets up the inequality $P(x) > 0$
• M1: Factorises the quadratic correctly
• A1: Correct final range

Common Mistake: Forgetting to reverse the inequality sign when dividing by a negative number. This is a very common error.

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Section B: Exponential and Logarithmic Functions (Questions 6–10)

Question 6 [3 marks]

Answer:

The graph of $y = e^x - 2$ is the graph of $y = e^x$ shifted down by 2 units.

Key features:

• y-intercept: When $x = 0$, $y = e^0 - 2 = 1 - 2 = -1$. So the y-intercept is $(0, -1)$.
• x-intercept: When $y = 0$, $e^x - 2 = 0 \Rightarrow e^x = 2 \Rightarrow x = \ln 2 \approx 0.693$. So the x-intercept is $(\ln 2, 0)$.
• Horizontal asymptote: As $x \to -\infty$, $e^x \to 0$, so $y \to -2$. The horizontal asymptote is $y = -2$.

Marking Notes:

• M1: Correct shape (exponential growth curve)
• A1: Correct y-intercept $(0, -1)$ and x-intercept $(\ln 2, 0)$
• A1: Correct horizontal asymptote $y = -2$

Common Mistake: Forgetting the horizontal asymptote or getting the y-intercept wrong. Remember that $e^0 = 1$.

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Question 7 [4 marks]

Answer: $x = 0.693$ or $x = 1.10$

Working:

Let $u = e^x$. Then $e^{2x} = (e^x)^2 = u^2$.

The equation becomes: $u^2 - 5u + 6 = 0$ $(u - 2)(u - 3) = 0$

So $u = 2$ or $u = 3$.

Since $u = e^x$:

• $e^x = 2 \Rightarrow x = \ln 2 \approx 0.693$
• $e^x = 3 \Rightarrow x = \ln 3 \approx 1.10$

Answer: $x = 0.693$ or $x = 1.10$ (3 s.f.)

Marking Notes:

• M1: Substitutes $u = e^x$ to form a quadratic in $u$
• M1: Factorises the quadratic correctly
• M1: Takes natural logs of both solutions
• A1: Both answers correct to 3 s.f.

Common Mistake: Trying to take logs of the original equation directly. The substitution method is essential here.

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Question 8 [4 marks]

Answer: $x = 2$

Working:

$\ln(x + 2) + \ln(x - 1) = \ln 4$

Using the product rule of logarithms: $\ln a + \ln b = \ln(ab)$

$\ln[(x + 2)(x - 1)] = \ln 4$

Since $\ln$ is a one-to-one function: $(x + 2)(x - 1) = 4$ $x^2 + x - 2 = 4$ $x^2 + x - 6 = 0$ $(x + 3)(x - 2) = 0$

So $x = -3$ or $x = 2$.

Check domain: We need $x + 2 > 0$ and $x - 1 > 0$, i.e., $x > -2$ and $x > 1$. So $x > 1$.

$x = -3$ does not satisfy $x > 1$, so it is rejected.

Answer: $x = 2$

Marking Notes:

• M1: Applies the product rule for logarithms
• M1: Equates the arguments (since $\ln$ is one-to-one)
• M1: Solves the quadratic and checks the domain
• A1: Correct final answer $x = 2$

Common Mistake: Forgetting to check the domain of the logarithmic function. $x = -3$ is an extraneous solution because $\ln(x+2)$ and $\ln(x-1)$ would be undefined (or have undefined arguments).

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Question 9 [2 marks]

Answer: The graph of $y = \ln x$ is transformed to $y = \ln(3x) + 1$ by:

1. A horizontal stretch by a scale factor of $\frac{1}{3}$ (i.e., a horizontal compression)
2. A vertical translation upwards by 1 unit

Working:

Starting from $y = \ln x$:

• $y = \ln(3x)$: This is a horizontal stretch by factor $\frac{1}{3}$ (the graph is compressed horizontally). This is because replacing $x$ with $3x$ compresses the graph horizontally.
• $y = \ln(3x) + 1$: This is a vertical translation upwards by 1 unit.

Marking Notes:

• A1: Correct horizontal transformation (stretch by factor $\frac{1}{3}$)
• A1: Correct vertical transformation (translation up by 1 unit)

Common Mistake: Confusing horizontal stretch factors. $y = f(3x)$ compresses the graph by factor $\frac{1}{3}$, not stretches by factor 3.

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Question 10 [3 marks]

Answer: $t = 6.9$ hours (1 d.p.)

Working:

We need $N(t) = 2000$: $500e^{0.2t} = 2000$ $e^{0.2t} = 4$ $0.2t = \ln 4$ $t = \frac{\ln 4}{0.2} = 5\ln 4 \approx 5(1.386) = 6.93$

Answer: $t = 6.9$ hours (1 d.p.)

Marking Notes:

• M1: Sets up the equation $500e^{0.2t} = 2000$
• M1: Takes natural logs of both sides
• A1: Correct answer to 1 d.p.

Common Mistake: Forgetting to divide by 500 first before taking logs. Always isolate the exponential term first.

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Section C: Graphing Techniques (Questions 11–15)

Question 11 [4 marks]

Answer:

The graph of $y = \frac{1}{x - 2} + 3$ is a rectangular hyperbola.

Key features:

• Vertical asymptote: $x = 2$ (where the denominator is zero)
• Horizontal asymptote: $y = 3$ (as $x \to \pm\infty$, $\frac{1}{x-2} \to 0$)
• x-intercept: Set $y = 0$: $\frac{1}{x-2} + 3 = 0 \Rightarrow \frac{1}{x-2} = -3 \Rightarrow x - 2 = -\frac{1}{3} \Rightarrow x = \frac{5}{3}$. So the x-intercept is $\left(\frac{5}{3}, 0\right)$.
• y-intercept: Set $x = 0$: $y = \frac{1}{0-2} + 3 = -\frac{1}{2} + 3 = \frac{5}{2}$. So the y-intercept is $\left(0, \frac{5}{2}\right)$.

Marking Notes:

• M1: Correct shape (two branches of a hyperbola)
• A1: Correct vertical asymptote $x = 2$ and horizontal asymptote $y = 3$
• A1: Correct x-intercept $\left(\frac{5}{3}, 0\right)$
• A1: Correct y-intercept $\left(0, \frac{5}{2}\right)$

Common Mistake: Getting the horizontal asymptote wrong. The $+3$ shifts the graph up, so the horizontal asymptote is $y = 3$, not $y = 0$.

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Question 12 [1 mark]

Answer: $(2, 8)$

Working:

The transformation $y = f(x) + 3$ is a vertical translation upwards by 3 units. This means every point on the graph moves up by 3 units.

The maximum point $(2, 5)$ becomes $(2, 5 + 3) = (2, 8)$.

Marking Notes:

• A1: Correct answer $(2, 8)$

Common Mistake: Adding 3 to the x-coordinate instead of the y-coordinate. Vertical translations affect y-coordinates only.

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Question 13 [4 marks]

Answer:

(a) $y = g(x - 1)$: This is a horizontal translation to the right by 1 unit. Every point on the graph moves right by 1 unit.

• Maximum: $(-2, 4) \to (-1, 4)$
• Minimum: $(1, -3) \to (2, -3)$
• x-intercepts: $(-4, 0) \to (-3, 0)$ and $(3, 0) \to (4, 0)$
• y-intercept: $(0, 2) \to (1, 2)$

(b) $y = -g(x)$: This is a reflection in the x-axis. Every point on the graph is reflected across the x-axis.

• Maximum: $(-2, 4) \to (-2, -4)$ (becomes a minimum)
• Minimum: $(1, -3) \to (1, 3)$ (becomes a maximum)
• x-intercepts: $(-4, 0) \to (-4, 0)$ and $(3, 0) \to (3, 0)$ (unchanged)
• y-intercept: $(0, 2) \to (0, -2)$

Marking Notes:

• (a) M1: Correct horizontal shift right by 1 unit, A1: Correct new coordinates
• (b) M1: Correct reflection in x-axis, A1: Correct new coordinates

Common Mistake: Confusing $g(x-1)$ with $g(x+1)$. $g(x-1)$ shifts right, $g(x+1)$ shifts left.

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Question 14 [2 marks]

Answer: Vertical asymptote: $x = 1$; Horizontal asymptote: $y = 2$

Working:

The transformation $y = h(x) + 2$ is a vertical translation upwards by 2 units.

• The vertical asymptote $x = 1$ is unaffected by vertical translations (it moves with the graph horizontally, but there's no horizontal shift here).
• The horizontal asymptote $y = 0$ moves up by 2 units to become $y = 2$.

Marking Notes:

• A1: Correct vertical asymptote $x = 1$
• A1: Correct horizontal asymptote $y = 2$

Common Mistake: Thinking the vertical asymptote also shifts. Vertical translations only affect horizontal asymptotes.

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Question 15 [3 marks]

Answer:

$y = x^2 - 4x + 3$

Factorising: $y = (x - 1)(x - 3)$

• x-intercepts: $x = 1$ and $x = 3$, i.e., $(1, 0)$ and $(3, 0)$
• y-intercept: $y = 3$, i.e., $(0, 3)$
• Turning point: Complete the square: $y = (x - 2)^2 - 1$. So the minimum point is $(2, -1)$.

Marking Notes:

• M1: Correct shape (U-shaped parabola opening upwards)
• A1: Correct x-intercepts $(1, 0)$ and $(3, 0)$, and y-intercept $(0, 3)$
• A1: Correct turning point $(2, -1)$

Common Mistake: Getting the sign of the turning point wrong. When completing the square, $y = (x-2)^2 - 1$ gives minimum at $(2, -1)$.

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Section D: Applications and Problem Solving (Questions 16–20)

Question 16 [3 marks]

Answer: Equilibrium price $p = 64$ dollars, equilibrium quantity $q = 18$ units.

Working:

At equilibrium, demand equals supply: $100 - 2q = 10 + 3q$ $100 - 10 = 3q + 2q$ $90 = 5q$ $q = 18$

Substitute $q = 18$ into either equation: $p = 100 - 2(18) = 100 - 36 = 64$

Answer: $q = 18$, $p = 64$

Marking Notes:

• M1: Equates the demand and supply functions
• M1: Solves for $q$
• A1: Correct equilibrium price $p = 64$

Common Mistake: Substituting back into the wrong equation or making arithmetic errors. Always check: $p = 10 + 3(18) = 10 + 54 = 64$. ✓

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Question 17 [3 marks]

Answer: (a) $y = (x - 3)^2 - 1$ (b) Minimum point: $(3, -1)$

Working:

(a) Complete the square: $y = x^2 - 6x + 8$ $y = (x^2 - 6x + 9) - 9 + 8$ $y = (x - 3)^2 - 1$

(b) The minimum point of $y = (x - a)^2 + b$ is $(a, b)$.

So the minimum point is $(3, -1)$.

Marking Notes:

• (a) M1: Correctly completes the square, A1: Correct form $(x - 3)^2 - 1$
• (b) A1: Correct minimum point $(3, -1)$

Common Mistake: Getting the sign wrong in the completed square form. $(x - 3)^2$ means the minimum is at $x = 3$, not $x = -3$.

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Question 18 [3 marks]

Answer: (a) $N(0) = 0$ fish (b) $N(10) = 787$ fish

Working:

(a) At $t = 0$: $N(0) = 2000(1 - e^0) = 2000(1 - 1) = 0$

(b) At $t = 10$: $N(10) = 2000(1 - e^{-0.05 \times 10}) = 2000(1 - e^{-0.5})$ $N(10) = 2000(1 - 0.6065) = 2000(0.3935) = 787$

Marking Notes:

• (a) A1: Correct answer $0$
• (b) M1: Correct substitution of $t = 10$, A1: Correct answer to nearest whole number

Common Mistake: Forgetting that $e^0 = 1$, so $N(0) = 0$. Also, rounding errors — keep full precision until the final step.

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Question 19 [5 marks]

Answer: (a) Stationary point: $(0, 1)$ (b) Minimum point

Working:

(a) Find the derivative: $y = e^x - x$ $\frac{dy}{dx} = e^x - 1$

At a stationary point, $\frac{dy}{dx} = 0$: $e^x - 1 = 0$ $e^x = 1$ $x = 0$

When $x = 0$: $y = e^0 - 0 = 1$

Stationary point: $(0, 1)$

(b) Find the second derivative: $\frac{d^2y}{dx^2} = e^x$

At $x = 0$: $\frac{d^2y}{dx^2} = e^0 = 1 > 0$

Since the second derivative is positive, the stationary point is a minimum.

Marking Notes:

• (a) M1: Correct differentiation, M1: Sets derivative to zero and solves, A1: Correct coordinates $(0, 1)$
• (b) M1: Correct second derivative, A1: Correct conclusion (minimum)

Common Mistake: Forgetting to find the y-coordinate of the stationary point. Also, confusing the first and second derivative tests.

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Question 20 [1 mark]

Answer: 2 real roots

Working:

The equation $f(x) = 0$ has real roots where the graph crosses the x-axis. From the graph, the curve crosses the x-axis at $x = -3$ and $x = 4$.

Therefore, there are 2 real roots.

Marking Notes:

• A1: Correct answer (2)

Common Mistake: Counting the turning points instead of the x-intercepts. The roots of $f(x) = 0$ are the x-intercepts of the graph.

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END OF ANSWER KEY