A Level H2 Mathematics Graphs Coordinate Geometry Quiz

A-Level Maths H2 Quiz - Graphs Coordinate Geometry (Answer Key)

1.
Vertical asymptote: Denominator is zero when $x + 1 = 0 \Rightarrow x = -1$.
Oblique asymptote: Perform long division or inspection.
$\frac{2x^2 - 5}{x + 1} = \frac{2x(x+1) - 2x - 5}{x+1} = 2x - \frac{2x+5}{x+1} = 2x - \frac{2(x+1)+3}{x+1} = 2x - 2 - \frac{3}{x+1}$.
As $x \to \infty$, $y \to 2x - 2$.
Answer: VA: $x = -1$, OA: $y = 2x - 2$. [2]

2.
Vertex at $2x - 4 = 0 \Rightarrow x = 2, y = 0$. Coordinates: $(2, 0)$.
$y$-intercept: $x=0 \Rightarrow y = |-4| = 4$. Coordinates: $(0, 4)$.
Graph is V-shaped, symmetric about $x=2$, passing through $(0,4)$ and $(2,0)$ and $(4,4)$.
Answer: Vertex $(2,0)$, $x$-int $(2,0)$. Sketch shows V-shape above x-axis. [2]

3.
$y = 2t \Rightarrow t = y/2$.
Substitute into $x$: $x = (y/2)^2 + 1 = \frac{y^2}{4} + 1$.
$4(x - 1) = y^2$ or $y^2 = 4x - 4$.
Answer: $y^2 = 4(x - 1)$. [2]

4.
$\frac{x-3}{x+2} - 1 > 0 \Rightarrow \frac{x - 3 - (x + 2)}{x + 2} > 0 \Rightarrow \frac{-5}{x + 2} > 0$.
Since numerator is negative, denominator must be negative.
$x + 2 < 0 \Rightarrow x < -2$.
Answer: $x < -2$. [2]

5.
Equation $|z - z_0| = r$ represents a circle with centre $z_0$ and radius $r$.
Here $z_0 = 2i$ (or $(0, 2)$) and $r = 3$.
Answer: Circle with centre $(0, 2)$ and radius $3$. [2]

6.
Asymptotes: Vertical when $x^2 - 4 = 0 \Rightarrow x = 2, x = -2$.
Horizontal as $x \to \infty, y \to 0 \Rightarrow y = 0$.
Graph: Even function. For $x=0, y = -1/4$. Between asymptotes, curve is below axis (u-shape inverted). Outside asymptotes, curve is above axis, approaching axes.
Answer: VA: $x = \pm 2$, HA: $y = 0$. [2]

7.
$\frac{dy}{dx} = 1 \cdot e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.
Stationary points when $\frac{dy}{dx} = 0 \Rightarrow 1 - x = 0 \Rightarrow x = 1$.
$y = 1 \cdot e^{-1} = \frac{1}{e}$.
Answer: $(1, \frac{1}{e})$. [2]

8.
Distance from centre $(0,0)$ to line $mx - y + c = 0$ must equal radius $5$.
Distance $d = \frac{|m(0) - 1(0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{m^2 + 1}}$.
$\frac{|c|}{\sqrt{m^2 + 1}} = 5 \Rightarrow |c| = 5\sqrt{m^2 + 1}$.
Square both sides: $c^2 = 25(m^2 + 1)$.
Answer: Shown. [2]

9.
Square both sides (valid as both sides non-negative):
$(2x - 1)^2 < (x + 3)^2$
$4x^2 - 4x + 1 < x^2 + 6x + 9$
$3x^2 - 10x - 8 < 0$
$(3x + 2)(x - 4) < 0$.
Critical values: $x = -2/3, x = 4$.
Parabola opens upward, so negative between roots.
Answer: $-\frac{2}{3} < x < 4$. [2]

10.
$y = \ln(x^2 + 1)$. $\frac{dy}{dx} = \frac{2x}{x^2 + 1}$.
At $x = 1$, $y = \ln(2)$. Gradient $m = \frac{2(1)}{1+1} = 1$.
Normal gradient $m_{\perp} = -1$.
Equation: $y - \ln 2 = -1(x - 1) \Rightarrow y = -x + 1 + \ln 2$.
Answer: $y = -x + 1 + \ln 2$. [2]

11.
(a) VA: $x = -2$. HA: $y = 3$ (coeff of $x$ / coeff of $x$).
$y$-int: $x=0 \Rightarrow y = -1/2$.
$x$-int: $y=0 \Rightarrow 3x-1=0 \Rightarrow x=1/3$.
Sketch: Hyperbola in top-right and bottom-left quadrants relative to asymptotes.
(b) $\frac{3x - 1}{x + 2} > 2 \Rightarrow \frac{3x - 1 - 2(x + 2)}{x + 2} > 0 \Rightarrow \frac{x - 5}{x + 2} > 0$.
Critical values: $5, -2$. Positive outside roots.
Answer: $x < -2$ or $x > 5$. [3]

12.
(a) $\cos \theta = x/3, \sin \theta = y/2$.
$\cos^2 \theta + \sin^2 \theta = 1 \Rightarrow (\frac{x}{3})^2 + (\frac{y}{2})^2 = 1 \Rightarrow \frac{x^2}{9} + \frac{y^2}{4} = 1$.
(b) Differentiate implicitly: $\frac{2x}{9} + \frac{2y}{4}\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{4x}{9y}$.
Set $-\frac{4x}{9y} = -\frac{2}{3} \Rightarrow \frac{2x}{3y} = 1 \Rightarrow y = \frac{2}{3}x$.
Sub into ellipse eq: $\frac{x^2}{9} + \frac{(2x/3)^2}{4} = 1 \Rightarrow \frac{x^2}{9} + \frac{4x^2/9}{4} = 1 \Rightarrow \frac{x^2}{9} + \frac{x^2}{9} = 1 \Rightarrow \frac{2x^2}{9} = 1$.
$x^2 = 4.5 \Rightarrow x = \pm \frac{3}{\sqrt{2}} = \pm \frac{3\sqrt{2}}{2}$.
$y = \frac{2}{3}(\pm \frac{3\sqrt{2}}{2}) = \pm \sqrt{2}$.
Answer: $(\frac{3\sqrt{2}}{2}, \sqrt{2})$ and $(-\frac{3\sqrt{2}}{2}, -\sqrt{2})$. [4]

13.
(i) Circle centre $(1, 1)$, radius $2$.
(ii) Ray from $(1, 1)$ at angle $\pi/4$ ($45^\circ$) to horizontal.
Region: Sector of the circle bounded by the ray and the horizontal line extending right from centre? No, argument is from positive real axis relative to centre.
$\arg(z - (1+i)) = \pi/4$ is a ray.
Inequality $0 \le \arg \dots \le \pi/4$ defines a wedge starting from horizontal right ($0$) to $45^\circ$ up.
Shade the sector within the circle between angle $0$ and $\pi/4$.
Answer: Diagram with circle centre $(1,1)$, shaded wedge from $0$ to $45^\circ$. [3]

14.
Oblique asymptote found by division: $\frac{x^2 + ax + b}{x - 1} = x + (a+1) + \frac{b + a + 1}{x - 1}$.
OA is $y = x + (a+1)$. Given $y = x + 3$.
So $a + 1 = 3 \Rightarrow a = 2$.
For VA at $x=1$, denominator is zero (already true). Numerator should not be zero at $x=1$ for simple pole, but actually the remainder term determines behavior.
Wait, if $x=1$ is VA, then $x-1$ does not cancel.
The constant term in quotient is $a+1$.
Let's check intercepts or specific points? No, just compare coefficients.
$x^2 + ax + b = (x-1)(x+3) + R$.
$(x-1)(x+3) = x^2 + 2x - 3$.
So $x^2 + ax + b = x^2 + 2x - 3 + R$.
Comparing coefficients of $x$: $a = 2$.
Constant term: $b = -3 + R$.
However, usually "oblique asymptote $y=x+3$" implies the polynomial part of the division is $x+3$.
Division: $(x^2+ax+b) \div (x-1)$.
$x(x-1) = x^2-x$. Subtract: $(a+1)x + b$.
$(a+1)(x-1) = (a+1)x - (a+1)$. Subtract: $b + a + 1$.
Quotient is $x + a + 1$.
So $a + 1 = 3 \Rightarrow a = 2$.
The remainder is $b + 2 + 1 = b + 3$.
Does $b$ affect the asymptote? No.
Is there another condition? "Vertical asymptote at $x=1$". This just means $x-1$ doesn't cancel.
If $b = -3$, numerator is $x^2+2x-3 = (x+3)(x-1)$, hole at $x=1$, no VA.
So $b \neq -3$.
Wait, did I miss a condition? Usually these questions fix $b$ via a point or intercept.
Re-reading: "Find the values of $a$ and $b$."
Is it possible the question implies the curve passes through a specific point? No.
Let's look at the structure. Maybe the asymptote intersection?
Usually, if only OA and VA are given, $b$ is not uniquely determined unless there's a constraint like "passes through origin" or similar.
However, in many exam contexts, if the numerator is quadratic and denominator linear, and OA is given, $a$ is fixed. $b$ can be any value such that $x=1$ is not a root.
Let's assume there is a typo in my generation or a standard convention.
Actually, often the question might say "The curve passes through $(0, -2)$" or similar.
Without extra info, $a=2$ and $b \neq -3$.
Let's adjust the question interpretation: Perhaps the "oblique asymptote" derivation includes the remainder going to 0.
Let's assume the question meant to provide a point, e.g., y-intercept is -1.
If $y$-int is $-1$: $b/-1 = -1 \Rightarrow b=1$.
Let's provide $a=2$ and state $b$ is any real number except $-3$.
Correction for Answer Key based on standard template: Often these questions come with a point. I will assume the question implies the simplest integer form or I missed a constraint.
Let's look at Q14 again. "Find the values". Plural.
If I assume the remainder is 0? No, then no VA.
Let's assume the question intended: $y = \frac{x^2+ax+b}{x-1}$ has OA $y=x+3$ and passes through $(2, 5)$.
$5 = \frac{4+2a+b}{1} \Rightarrow 2a+b=1$. With $a=2, 4+b=1 \Rightarrow b=-3$. But $b=-3$ removes VA.
Let's try $(0, 1)$. $1 = \frac{b}{-1} \Rightarrow b=-1$.
I will provide $a=2$ and note that $b$ cannot be $-3$. For the sake of a definitive answer key, I will assume a standard constraint like "y-intercept is 2" was intended but omitted in the prompt generation, OR I will simply solve for $a$ and leave $b$ as a condition.
Self-Correction: I will state $a=2$ and $b \in \mathbb{R} \setminus \{-3\}$.
Answer: $a = 2$, $b \neq -3$. [3]

15.
$y = x^2 - 4x + 3 = (x-1)(x-3)$. Roots at $1, 3$. Vertex at $x=2, y = 4-8+3 = -1$.
Absolute value reflects the part below x-axis.
Vertex becomes $(2, 1)$.
Intercepts: $(1,0), (3,0), (0,3)$.
Turning points: $(2,1)$ (local max), $(1,0)$ and $(3,0)$ (local minima).
Answer: TP: $(2,1), (1,0), (3,0)$. Y-int: $(0,3)$. W-shape. [2]

16.
(a) Vertex $(x, 12-x^2)$. Base width $2x$. Height $12-x^2$.
$A(x) = 2x(12 - x^2) = 24x - 2x^3$.
(b) $\frac{dA}{dx} = 24 - 6x^2$.
Set to 0: $6x^2 = 24 \Rightarrow x^2 = 4 \Rightarrow x = 2$ (since $x>0$).
Max Area $A(2) = 2(2)(12 - 4) = 4(8) = 32$.
Answer: (a) $A = 24x - 2x^3$, (b) $32$. [3]

17.
(a) $y=1/x$ hyperbola, $y=x^2-2$ parabola.
(b) Intersection: $\frac{1}{x} = x^2 - 2 \Rightarrow 1 = x^3 - 2x \Rightarrow x^3 - 2x - 1 = 0$.
(c) Check integer roots. $x=-1: -1+2-1=0$. So $(x+1)$ is factor.
$x^3 - 2x - 1 = (x+1)(x^2 - x - 1) = 0$.
$x = -1$ or $x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$.
Answer: $x = -1, \frac{1 + \sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2}$. [3]

18.
(a) Locus of points equidistant from $4$ (point $(4,0)$) and $-2i$ (point $(0,-2)$). Perpendicular bisector of segment joining $(4,0)$ and $(0,-2)$.
(b) Midpoint: $(2, -1)$. Gradient of segment: $\frac{-2-0}{0-4} = \frac{-2}{-4} = \frac{1}{2}$.
Gradient of bisector: $-2$.
Eq: $y - (-1) = -2(x - 2) \Rightarrow y + 1 = -2x + 4 \Rightarrow y = -2x + 3$ or $2x + y - 3 = 0$.
(c) Min $|z|$ is distance from origin to line $2x + y - 3 = 0$.
$d = \frac{|2(0) + 1(0) - 3|}{\sqrt{2^2 + 1^2}} = \frac{3}{\sqrt{5}}$.
Answer: (a) Perp bisector, (b) $2x+y=3$, (c) $\frac{3}{\sqrt{5}}$. [3]

19.
(a) $x^2 = t^2 + 2 + 1/t^2$. $y^2 = t^2 - 2 + 1/t^2$.
$x^2 - y^2 = (t^2 + 2 + 1/t^2) - (t^2 - 2 + 1/t^2) = 4$.
(b) Hyperbola. Since $t>0$, $x = t+1/t \ge 2$ (AM-GM). Right branch only.
(c) $t=2 \Rightarrow x = 2.5, y = 1.5$.
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. $dx/dt = 1 - 1/t^2$, $dy/dt = 1 + 1/t^2$.
At $t=2$: $dx/dt = 1 - 1/4 = 3/4$. $dy/dt = 1 + 1/4 = 5/4$.
$m = \frac{5/4}{3/4} = \frac{5}{3}$.
Eq: $y - 1.5 = \frac{5}{3}(x - 2.5)$.
$y - \frac{3}{2} = \frac{5}{3}(x - \frac{5}{2})$.
$6y - 9 = 10x - 25 \Rightarrow 10x - 6y - 16 = 0 \Rightarrow 5x - 3y - 8 = 0$.
Answer: (a) Shown, (b) Right branch hyperbola, (c) $5x - 3y - 8 = 0$. [4]

20.
(a) $g(x) = (x-2)^2 + 1$. Vertex at $x=2$. For 1-1, domain must be one side of vertex. Smallest $k=2$.
(b) $y = (x-2)^2 + 1 \Rightarrow y - 1 = (x-2)^2 \Rightarrow x - 2 = \sqrt{y-1}$ (since $x \ge 2$).
$x = 2 + \sqrt{y-1}$.
$g^{-1}(x) = 2 + \sqrt{x-1}$.
Domain of $g^{-1}$ = Range of $g$. Min $g(2)=1$. Domain: $x \ge 1$.
(c) Sketch: $g(x)$ is right half of parabola vertex $(2,1)$. $g^{-1}(x)$ is upper half of sideways parabola vertex $(1,2)$. Intersect on $y=x$.
Answer: (a) $k=2$, (b) $g^{-1}(x) = 2+\sqrt{x-1}, x \ge 1$, (c) Sketch. [4]