Questions
Consider the function \(f(x)=\frac{4x+2}{x-2}, x\neq 2\)
(a) Sketch the graph of \(y=f(x)\). On your sketch, indicate the values of any axis intercepts and label any asymptotes with their equations.
(b) Write down the range of f.
Consider the function \(g(x)=x^{2}+bx+c\). The graph of g has an axis of symmetry at \(x=2\).
The two roots of \(g(x)=0\) are \(-\frac{1}{2}\) and p, where \(p\in \mathbb{Q}\).
(c) Show that \(p=\frac{9}{2}\).
(d) Find the value of b and the value of c.
(e) Find the y-coordinate of the vertex of the graph of \(y=g(x)\).
(f) Find the number of solutions of the equation \(f(x)=g(x)\).
▶️Answer/Explanation
Detailed solution
(a) Sketch the Graph of \( y = f(x) \)
We’re sketching \( y = \frac{4x + 2}{x – 2} \), where \( x \neq 2 \), and need to identify axis intercepts and asymptotes. This is a rational function, so expect vertical and horizontal asymptotes due to the division.
Vertical Asymptote
The denominator \( x – 2 = 0 \) at \( x = 2 \), and since the numerator \( 4x + 2 \) isn’t zero there (\( 4 \cdot 2 + 2 = 10 \)), there’s a vertical asymptote at:
\[
x = 2
\]
As \( x \to 2^- \), \( x – 2 \to 0^- \), and the numerator is positive, so \( y \to -\infty \). As \( x \to 2^+ \), \( x – 2 \to 0^+ \), so \( y \to +\infty \).
Horizontal Asymptote
For large \( |x| \), compare leading coefficients of the numerator (4x) and denominator (x):
\[
y \approx \frac{4x}{x} = 4
\]
So, the horizontal asymptote is:
\[
y = 4
\]
Confirm with limits: As \( x \to \infty \), \( y = \frac{4x + 2}{x – 2} = \frac{4 + \frac{2}{x}}{1 – \frac{2}{x}} \to \frac{4}{1} = 4 \), and similarly as \( x \to -\infty \).
x-Intercept
Set \( y = 0 \):
\[
\frac{4x + 2}{x – 2} = 0
\]
Numerator must be zero (denominator \( \neq 0 \)):
\[
4x + 2 = 0 \quad \Rightarrow \quad 4x = -2 \quad \Rightarrow \quad x = -\frac{1}{2}
\]
Check: \( x – 2 = -\frac{1}{2} – 2 = -\frac{5}{2} \neq 0 \). So, x-intercept is \( \left(-\frac{1}{2}, 0\right) \).
y-Intercept
Set \( x = 0 \):
\[
y = \frac{4 \cdot 0 + 2}{0 – 2} = \frac{2}{-2} = -1
\]
y-intercept is \( (0, -1) \).
Sketch Details
– Vertical asymptote at \( x = 2 \).
– Horizontal asymptote at \( y = 4 \).
– Intercepts: \( \left(-\frac{1}{2}, 0\right) \) and \( (0, -1) \).
– Behavior: Left of \( x = 2 \), test \( x = 1 \): \( y = \frac{4 + 2}{1 – 2} = \frac{6}{-1} = -6 \) (below y = 4). Right of \( x = 2 \), test \( x = 3 \): \( y = \frac{12 + 2}{3 – 2} = 14 \) (above y = 4).
The graph has two branches as shown below : one approaching \( y = 4 \) from below as \( x \to -\infty \), crossing the x-axis at \( -\frac{1}{2} \), dipping to \( -1 \) at the y-axis, then plunging to \(-\infty\) as \( x \to 2^- \). The other rises from \( +\infty \) at \( x = 2^+ \) toward \( y = 4 \).
(b) Write Down the Range of \( f \)
Since \( f(x) = \frac{4x + 2}{x – 2} \) is continuous except at \( x = 2 \), the range excludes values not achieved. Solve \( y = f(x) \) for \( x \):
\[
y = \frac{4x + 2}{x – 2}
\]
\[
y (x – 2) = 4x + 2
\]
\[
yx – 2y = 4x + 2
\]
\[
yx – 4x = 2y + 2
\]
\[
x (y – 4) = 2y + 2
\]
\[
x = \frac{2y + 2}{y – 4}, \quad y \neq 4
\]
\( y = 4 \) makes the denominator zero, undefined (matches the horizontal asymptote).
\( x = 2 \) (undefined point) gives: \( 2 = \frac{2y + 2}{y – 4} \), so \( 2y – 8 = 2y + 2 \), \( -8 = 2 \), a contradiction—no \( y \) value maps to \( x = 2 \).
As \( x \) varies over \( (-\infty, 2) \cup (2, \infty) \), \( y \) covers all reals except \( y = 4 \) (the asymptote value never reached). Range:
\[
\mathbb{R} \setminus \{4\}
\]
(c) Show that \( p = \frac{9}{2} \)
For \( g(x) = x^2 + bx + c \), the axis of symmetry is at \( x = 2 \), and roots are \( -\frac{1}{2} \) and \( p \). For a quadratic \( ax^2 + bx + c \), the axis of symmetry is \( x = -\frac{b}{2a} \). Here, \( a = 1 \), so:
\[
-\frac{b}{2 \cdot 1} = 2 \quad \Rightarrow \quad -\frac{b}{2} = 2 \quad \Rightarrow \quad b = -4
\]
Roots are \( -\frac{1}{2} \) and \( p \). Sum of roots:
\[
-\frac{1}{2} + p = -\frac{b}{a} = -(-4) = 4
\]
\[
p – \frac{1}{2} = 4 \quad \Rightarrow \quad p = 4 + \frac{1}{2} = \frac{8}{2} + \frac{1}{2} = \frac{9}{2}
\]
So, \( p = \frac{9}{2} \), a rational number, as required.
(d) Find \( b \) and \( c \)
From (c), \( b = -4 \). Product of roots gives \( c \):
\[
\left(-\frac{1}{2}\right) \cdot \frac{9}{2} = \frac{c}{a} = c \quad (a = 1)
\]
\[
-\frac{9}{4} = c
\]
Thus, \( b = -4 \), \( c = -\frac{9}{4} \). Check: \( g(x) = x^2 – 4x – \frac{9}{4} \), roots via quadratic formula:
\[
x = \frac{4 \pm \sqrt{16 – 4 \cdot 1 \cdot (-\frac{9}{4})}}{2} = \frac{4 \pm \sqrt{16 + 9}}{2} = \frac{4 \pm \sqrt{25}}{2} = \frac{4 \pm 5}{2}
\]
\[
x = \frac{9}{2}, \quad x = -\frac{1}{2}
\]
(e) Find the y-Coordinate of the Vertex
Vertex x-coordinate is the axis of symmetry, \( x = 2 \). Compute:
\[
g(2) = 2^2 – 4 \cdot 2 – \frac{9}{4} = 4 – 8 – \frac{9}{4} = -4 – \frac{9}{4} = -\frac{16}{4} – \frac{9}{4} = -\frac{25}{4}
\]
y-coordinate is \(-\frac{25}{4}\).
(f) Find the Number of Solutions to \( f(x) = g(x) \)
Set:
\[
\frac{4x + 2}{x – 2} = x^2 – 4x – \frac{9}{4}
\]
Multiply through by \( x – 2 \) (\( x \neq 2 \)):
\[
4x + 2 = \left(x^2 – 4x – \frac{9}{4}\right) (x – 2)
\]
Right side:
\[
x^2 (x – 2) – 4x (x – 2) – \frac{9}{4} (x – 2) = x^3 – 2x^2 – 4x^2 + 8x – \frac{9}{4} x + \frac{18}{4} = x^3 – 6x^2 + \frac{32}{4} x + \frac{18}{4} = x^3 – 6x^2 + 8x + \frac{9}{2}
\]
Equation:
\[
4x + 2 = x^3 – 6x^2 + 8x + \frac{9}{2}
\]
\[
0 = x^3 – 6x^2 + 8x + \frac{9}{2} – 4x – 2
\]
\[
0 = x^3 – 6x^2 + 4x + \frac{9}{2} – \frac{4}{2} = x^3 – 6x^2 + 4x + \frac{5}{2}
\]
Solve the cubic \( x^3 – 6x^2 + 4x + \frac{5}{2} = 0 \). Test roots:
\( x = -\frac{1}{2} \): \( -\frac{1}{8} – 6 \cdot \frac{1}{4} + 4 \cdot (-\frac{1}{2}) + \frac{5}{2} = -\frac{1}{8} – \frac{6}{4} – 2 + \frac{5}{2} = -\frac{1}{8} – \frac{12}{8} – \frac{16}{8} + \frac{20}{8} = \frac{-1 – 12 – 16 + 20}{8} = -\frac{9}{8} \neq 0 \).
Use the Rational Root Theorem: Possible roots like \( \pm \frac{5}{2} \):
\( x = -\frac{5}{2} \): \( -\frac{125}{8} – 6 \cdot \frac{25}{4} + 4 \cdot (-\frac{5}{2}) + \frac{5}{2} = -\frac{125}{8} – \frac{150}{4} – 10 + \frac{5}{2} = -\frac{125}{8} – \frac{300}{8} – \frac{80}{8} + \frac{20}{8} = \frac{-125 – 300 – 80 + 20}{8} = -\frac{485}{8} \neq 0 \).
Factor or analyze intersections graphically:
\( g(x) \) is a parabola opening upward, vertex at \( (2, -\frac{25}{4}) \), roots at \( -\frac{1}{2} \) and \( \frac{9}{2} \).
\( f(x) \) has a discontinuity at \( x = 2 \).
Discriminant of the cubic (or intersection count via graphing):
– Define \( h(x) = f(x) – g(x) = 0 \). As \( x \to 2^- \), \( f(x) \to -\infty \), \( g(2) = -\frac{25}{4} \), so \( h(x) \to -\infty \). As \( x \to 2^+ \), \( h(x) \to +\infty \). Between roots of \( g(x) \), \( g \) is negative, \( f \) crosses from negative to positive.
Typically, a cubic has 1 or 3 real roots. Testing shows no obvious rational roots easily, but graphically, \( f \) and \( g \) intersect thrice , suggesting 3 solutions as observed in the graph below, considering the asymptote behavior.
Ans:
(a)
vertical asymptote \(x=2)\) sketched and labelled with correct equation
horizontal asymptote \(y=4\) sketched and labelled with correct equation
For an approximate rational function shape:
labelled intercepts \(-\frac{1}{2}\) on x-axis, \(-1\) on y-axis
two branches in correct opposite quadrants with correct asymptotic behaviour
(b) \(y\neq 4\) (or equivalent)
(c) 
(d) METHOD 1
attempt to substitute both roots to form a quadratic
EITHER
OR
METHOD 2
\(-\frac{b}{2}=2\) OR \(4+b=0\Rightarrow b=-4\)
attempt to form a valid equation to find c using their b
METHOD 3
attempt to form two valid equations in b and c
METHOD 4
attempt to write \(g(x)\) in the form \((x-h)^{2}+k\) and substitute x, h and \(g(x)\)
(e) attempt to substitute \(x=2\) into their \(g(x)\) OR complete the square on their \(g(x)\) (may be seen in part (d))
\(y=-\frac{25}{4}\)
(f)
both graphs sketched on same axes and identifying points of intersection
3 solutions