Chapter 2: Introduction to Linear Polynomials
Chapter 2: Introduction to Linear Polynomials
Exercise
Question
Question: 13
25 / 26
25 / 26
Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
(i) The graph of p(x) passes through the points (2, 3) and (6, 11).
(ii) The graph of q(x) passes through the point (4, -1).
(iii) The graph of q(x) is parallel to the graph of p(x).
Find the polynomials p(x) and q(x). Also, find the coordinates of the points where lines meet the x-axis.
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Answer
1
Use the point (2, 3) to form Equation 1 for p(x).
\text{${color:#2E2EFF}3{/color} = a × {color:#2E2EFF}2{/color} + b$}
⇒
\text{ $3 = 2a + b$ $\quad$ ${color:#6D28D9}[Equation 1]{/color}$}
2
Use the point (6, 11) to form Equation 2.
\text{${color:#2E2EFF}11{/color} = a × {color:#2E2EFF}6{/color} + b$}
⇒
\text{ $11 = 6a + b$ $\quad$ ${color:#6D28D9}[Equation 2]{/color}$}
3
Subtract Equation 1 from Equation 2 to find a.
$11 - 3 = (6a + b) - (2a + b)$
⇒
$8 = (6a - 2a) + (b - b)$
⇒
$8 = 4a$
⇒
\text{ $a = 2$ $\quad$ ${color:#625B5F}[Divide both sides by 4]{/color}$}
4
Substitute the value of a into Equation 1 to find b.
\text{$3 = 2 × {color:#2E2EFF}2{/color} + b$}
⇒
$3 = 4 + b$
⇒
$b = 3 - 4$
⇒
$b = -1$
5
Substitute the values of a and b into p(x) = ax + b to write the polynomial.
Answer: $p(x) = 2x - 1$
6
Since the graphs of p(x) and q(x) are parallel, their slopes are equal. Hence, c = a = 2.
$q(x) = cx + d$
⇒
\text{ $q(x) = {color:#2E2EFF}2{/color}x + d$}
7
Use the point (4, -1) to find d.
\text{${color:#2E2EFF}-1{/color} = 2 × {color:#2E2EFF}4{/color} + d$}
⇒
$-1 = 8 + d$
⇒
$d = -1 - 8$
⇒
$d = -9$
8
Substitute the values of c and d into q(x) = cx + d to write the polynomial.
Answer: $q(x) = 2x - 9$
9
Set p(x) = 0 to find its x-intercept. At the x-intercept, y = 0.
$0 = 2x - 1$
⇒
$2x = 1$
⇒
\text{ $x = \dfrac{1}{2}$}
10
Set q(x) = 0 to find its x-intercept.
$0 = 2x - 9$
⇒
$2x = 9$
⇒
\text{ $x = \dfrac{9}{2}$}
11
Write the coordinates of both x-intercepts.
\textbf{Answer: The graph of $p(x)$ cuts the $x$-axis at $(\dfrac{1}{2}, 0)$, and the graph of $q(x)$ cuts the $x$-axis at $(\dfrac{9}{2}, 0)$}