Question
Consider the curve $y = \frac{x-4}{ax^2 + bx + c}$, where $a$, $b$, and $c$ are non-zero constants.
The curve has a local minimum point at $(2, 1)$ and a vertical asymptote with equation $x = 1$.
Find the values of $a$, $b$, and $c$.
▶️Answer/Explanation
Detailed Solution
The curve is defined as:
\[
y = \frac{x – 4}{a x^2 + b x + c}
\]
Given
1. The curve has a local minimum at the point \( (2, 1) \).
2. The curve has a vertical asymptote at \( x = 1 \).
We will use these conditions to determine \( a \), \( b \), and \( c \), which are non-zero constants.
Step 1: Use the vertical asymptote to find a relationship between \( a \), \( b \), and \( c \)
A vertical asymptote occurs where the denominator of a rational function is zero, as the function approaches infinity or negative infinity. The denominator of our function is:
\[
a x^2 + b x + c
\]
The vertical asymptote is at \( x = 1 \), so the denominator must be zero when \( x = 1 \):
\[
a (1)^2 + b (1) + c = 0
\]
\[
a + b + c = 0 \tag{1}
\]
This gives us our first equation relating \( a \), \( b \), and \( c \).
Step 2: Use the point \( (2, 1) \) to get another equation
Since the point \( (2, 1) \) lies on the curve, substituting \( x = 2 \) and \( y = 1 \) into the equation of the curve should satisfy it:
\[
y = \frac{x – 4}{a x^2 + b x + c}
\]
\[
1 = \frac{2 – 4}{a (2)^2 + b (2) + c}
\]
\[
1 = \frac{-2}{4a + 2b + c}
\]
Multiply through by the denominator:
\[
4a + 2b + c = -2 \tag{2}
\]
This is our second equation.
Step 3: Use the local minimum condition at \( (2, 1) \)
The point \( (2, 1) \) is a local minimum, which means the derivative \( \frac{dy}{dx} = 0 \) at \( x = 2 \), and the second derivative \( \frac{d^2 y}{dx^2} > 0 \) at \( x = 2 \) (to confirm it’s a minimum). Let’s start by finding the first derivative.
The function is a quotient, so we use the quotient rule. Let:
– Numerator: \( u = x – 4 \), so \( \frac{du}{dx} = 1 \)
– Denominator: \( v = a x^2 + b x + c \), so \( \frac{dv}{dx} = 2a x + b \)
The quotient rule states:
\[
\frac{dy}{dx} = \frac{\frac{du}{dx} v – u \frac{dv}{dx}}{v^2}
\]
\[
\frac{dy}{dx} = \frac{(1)(a x^2 + b x + c) – (x – 4)(2a x + b)}{(a x^2 + b x + c)^2}
\]
\[
= \frac{a x^2 + b x + c – (x – 4)(2a x + b)}{(a x^2 + b x + c)^2}
\]
Expand the numerator:
\[
(x – 4)(2a x + b) = x (2a x + b) – 4 (2a x + b) = 2a x^2 + b x – 8a x – 4b
\]
\[
= 2a x^2 + (b – 8a) x – 4b
\]
So the numerator becomes:
\[
a x^2 + b x + c – (2a x^2 + (b – 8a) x – 4b) = a x^2 + b x + c – 2a x^2 – b x + 8a x + 4b
\]
\[
= (a – 2a) x^2 + (b – b + 8a) x + (c + 4b) = -a x^2 + 8a x + (c + 4b)
\]
Thus:
\[
\frac{dy}{dx} = \frac{-a x^2 + 8a x + c + 4b}{(a x^2 + b x + c)^2}
\]
At a local minimum, \( \frac{dy}{dx} = 0 \), so the numerator must be zero:
\[
-a x^2 + 8a x + c + 4b = 0
\]
Substitute \( x = 2 \):
\[
-a (2)^2 + 8a (2) + c + 4b = 0
\]
\[
-4a + 16a + c + 4b = 0
\]
\[
12a + c + 4b = 0 \tag{3}
\]
This is our third equation.Step 4: Solve the system of equations
We now have three equations:
1. \( a + b + c = 0 \)
2. \( 4a + 2b + c = -2 \)
3. \( 12a + 4b + c = 0 \)
**Solve equations (1) and (2):**
Subtract (1) from (2):
\[
(4a + 2b + c) – (a + b + c) = -2 – 0
\]
\[
3a + b = -2 \tag{4}
\]
Solve equations (2) and (3):
Subtract (2) from (3):
\[
(12a + 4b + c) – (4a + 2b + c) = 0 – (-2)
\]
\[
8a + 2b = 2
\]
\[
4a + b = 1 \tag{5}
\]
Solve equations (4) and (5):
Subtract (4) from (5):
\[
(4a + b) – (3a + b) = 1 – (-2)
\]
\[
a = 3
\]
Substitute \( a = 3 \) into (4):
\[
3(3) + b = -2
\]
\[
9 + b = -2
\]
\[
b = -11
\]
Substitute \( a = 3 \) and \( b = -11 \) into (1):
\[
3 + (-11) + c = 0
\]
\[
-8 + c = 0
\]
\[
c = 8
\]
So, we have:
\[
a = 3, \quad b = -11, \quad c = 8
\]
Step 5: Verify the solution
Let’s ensure these values satisfy all conditions.
Check the vertical asymptote:
\[
a x^2 + b x + c = 3 x^2 – 11 x + 8
\]
At \( x = 1 \):
\[
3(1)^2 – 11(1) + 8 = 3 – 11 + 8 = 0
\]
The denominator is zero, confirming the vertical asymptote at \( x = 1 \).
Check the point \( (2, 1) \):
\[
y = \frac{x – 4}{3 x^2 – 11 x + 8}
\]
At \( x = 2 \):
\[
y = \frac{2 – 4}{3(2)^2 – 11(2) + 8} = \frac{-2}{3(4) – 22 + 8} = \frac{-2}{12 – 22 + 8} = \frac{-2}{-2} = 1
\]
The point \( (2, 1) \) lies on the curve.
Check the local minimum:
We already used \( \frac{dy}{dx} = 0 \) at \( x = 2 \) to get our third equation, so let’s confirm it’s a minimum by checking the second derivative or the sign of the first derivative around \( x = 2 \). For simplicity, let’s recompute \( \frac{dy}{dx} \) with our values:
\[
y = \frac{x – 4}{3 x^2 – 11 x + 8}
\]
\[
\frac{dy}{dx} = \frac{(1)(3 x^2 – 11 x + 8) – (x – 4)(6 x – 11)}{(3 x^2 – 11 x + 8)^2}
\]
\[
= \frac{3 x^2 – 11 x + 8 – (6 x^2 – 11 x – 24 x + 44)}{(3 x^2 – 11 x + 8)^2}
\]
\[
= \frac{3 x^2 – 11 x + 8 – 6 x^2 + 35 x – 44}{(3 x^2 – 11 x + 8)^2}
\]
\[
= \frac{-3 x^2 + 24 x – 36}{(3 x^2 – 11 x + 8)^2}
\]
At \( x = 2 \), the numerator:
\[
-3(2)^2 + 24(2) – 36 = -12 + 48 – 36 = 0
\]
This confirms the critical point. To verify it’s a minimum, we could compute the second derivative, but a quicker check is to evaluate the sign of \( \frac{dy}{dx} \) around \( x = 2 \):
At \( x = 1.5 \):
\[
-3(1.5)^2 + 24(1.5) – 36 = -3(2.25) + 36 – 36 = -6.75 < 0
\]
At \( x = 3 \):
\[
-3(3)^2 + 24(3) – 36 = -27 + 72 – 36 = 9 > 0
\]
The derivative changes from negative to positive, confirming a local minimum at \( x = 2 \).
……………………..Markscheme…………………………
Solution: –
recognizes that $x = 1 \Rightarrow ax^2 + bx + c = 0$
$a + b + c = 0$ (seen anywhere)
passes through (2, 1) so:
$1 = \frac{2-4}{4a+2b+c} (4a + 2b + c = -2)$ (seen anywhere)
local minimum point at (2, 1) so:
attempts to find $\frac{dy}{dx}$ using quotient or product rule
$\frac{dy}{dx} = \frac{(ax^2 + bx + c) – (x – 4)(2ax + b)}{(ax^2 + bx + c)^2}$
substitutes $x = 2$ into the numerator of their $\frac{dy}{dx} = 0$
$(4a + 2b + c) – (2 – 4)(4a + b) = 0 (\frac{(4a + 2b + c) – (2 – 4)(4a + b)}{(4a + 2b + c)^2} = 0)$
$(12a + 4b + c = 0)$
attempts to solve their 3 linear equations in $a$, $b$, and $c$
$a = 3, b = -11$ and $c = 8$