Quadratic Function တစ်ခု၏ Standard Form ကို $y=ax^2+bx+c, a\ne 0$ ဟူ၍လည်းကောင်း ၊ Vertex Form ကို $y=a(x-h)^2+k a\ne 0$ ဟူ၍လည်းကောင်း၊ Grade (10) သင်ရိုးသစ် သင်ခန်းစာတွင် သိရှိပြီး ဖြစ်သည်။ Standard Form ကို $y=ax^2+bx+c$ နှင့် Vertex Form $y=a(x-h)^2+k $ ကိုလည်း ပုံစံ တစ်ခုမှ တစ်ခုသို့ အပြန်အလှန် ပြောင်းနိုင်ကြောင်း သိရှိခဲ့ပြီး ဖြစ်သည်။
$y=ax^2+bx+c, a\ne 0$ မှ $y=a(x-h)^2+k $ သို့ ပြောင်းလျှင် $h=-\dfrac{b}{2a}$ နှင့် $k=\dfrac{-b^2+4ac}{4a}$ ဖြစ်ကြောင်းသိရမည်။
Quadractic Function တစ်ခု၏ graph ကို parabola ဟု ခေါပြီး $a>0$ (positive) ဖြစ်လျှင် open upward ဖြစ်ကြောင်းနှင့် $a<0$ (negative) ဖြစ်လျှင် open downward ဖြစ်ကြောင်းသိရမည်။
Vertex Form $y=a(x-h)^2+k $ တွင် Parabola ၏ vertex (or turning point) သည် $(h,k)$ ဖြစ်ပြီး open upward ပုံစံတွင် $(h,k)$ သည် graph ပေါ်ရှိ အနိမ့်ဆုံးအမှတ် (minimum point) ဖြစ်ပြီး open downward ပုံစံတွင် $(h,k)$ သည် graph ပေါ်ရှိ အမြင့်ဆုံးအမှတ် (maximum point) ဖြစ်ကြောင်းသိရမည်။
Standard Form $y=ax^2+bx+c$ ပုံစံမှ $b^2-4ac$ တန်ဖိုးကို Quadratic Function တစ်ခု၏ Discriminant ဟုခေါ်ပြီး
- Discriminant = $b^2-4ac>0$ ဖြစ်လျှင် Parabola သည် $x$ ဝင်ရိုးကို အမှတ်နှစ်ခု၌ ဖြတ်သည်။ လက်ဝဲဘက် ဖြတ်မှတ်သည် $\left(\dfrac{-b-\sqrt{b^2-4ac}}{2a},0\right)$ ဖြစ်ပြီး လက်ယာဘက် ဖြတ်မှတ်သည် ဖြတ်မှတ်သည် $\left(\dfrac{-b+\sqrt{b^2-4ac}}{2a},0\right)$ ဖြစ်သည်။
- Discriminant = $b^2-4ac=0$ ဖြစ်လျှင် Parabola သည် $x$ ဝင်ရိုးကို အမှတ်တစ်ခု၌ ဖြတ်ပြီး ထိုအမှတ်သည် vertex ဖြစ်သည်။
- Discriminant = $b^2-4ac<0$ ဖြစ်လျှင် Parabola သည် $x$ ဝင်ရိုးကို လုံးဝမဖြတ်တော့ပါ။
$ax^2+bx+c =0$ ညီမျှင်ခြင်းကို Quadractic Equation ဟုခေါ်သည်။ ထိုညီမျှခြင်းကို ပြေလည်စေသော $x$ တန်ဖိုးများကို roots ဟုခေါ်သည်။ အဆိုပါ roots များသည် Parabola က $x$ ဝင်ရိုးကို ဖြတ်သော $x$-intercept များနှင့် အတူတူပင်ဖြစ်သည်။ ထို့ကြောင့်
- Discriminant = $b^2-4ac>0$ ဖြစ်လျှင် 2 roots (ညီမျှင်ခြင်းကို ပြေလည်စေသော အဖြေနှစ်ခုရှိသည်။) ၎င်းတို့မှာ $x_1=\dfrac{-b-\sqrt{b^2-4ac}}{2a}$ နှင့် $x_2=\dfrac{-b+\sqrt{b^2-4ac}}{2a}$ တို့ ဖြစ်သည်။
- Discriminant = $b^2-4ac=0$ ဖြစ်လျှင် Pဖြစ်လျှင် 1 root or repeated roots (ညီမျှင်ခြင်းကို ပြေလည်စေသော အဖြေတစ်ခုသာရှိသည်။) ၎င်းမှာ $x_1=x_2=\dfrac{-b}{2a}$ ဖြစ်သည်။
- Discriminant = $b^2-4ac<0$ ဖြစ်လျှင် no root (ညီမျှင်ခြင်းကို ပြေလည်စေသော အဖြေမရှိပါ)
Discriminant = $b^2-4ac>0$ ဖြစ်လျှင် Quadratic Function တစ်ခုကို $ax^2+bx+c=a(x-p)(x-q)$ ဟု ဖေါ်ပြနိုင်ပြီး $p=\dfrac{-b-\sqrt{b^2-4ac}}{2a}$ နှင့် $q=\dfrac{-b+\sqrt{b^2-4ac}}{2a}$ ဖြစ်သည်။ အဆိုပါအခြေအနေတွင်
- $p+q \text{ (sum of roots)} = -\dfrac{b}{a} $ ဖြစ်သည်။
- $pq \text{ (product of roots)} = \dfrac{c}{a} $ ဖြစ်သည်။
အောက်ပါပုစ္ဆာများသည် IGCSE (AS Level မှ) လက်တွေ့နယ်ပယ်တွင် တွေ့ရလေ့ရှိသည့် Quadractic Function ဆိုင်ရာ Real Life Problems များဖြစ်ပါသည်။ Grade 10 သင်ရိုးသစ် Quadratic Functions သင်ခန်းစာကို ကောင်းစွာ နားလည်သဘောပေါက်လျှင် အလွယ်တကူတွက်နိုင်သော မေးခွန်းများ ဖြစ်သည်။
Question 1
The diagram shows a section of a suspension bridge carrying a road over water. The height of the cables above water level in metres can be modelled by the function $h(x)=0.00012 x^{2}+200$, where $x$ is the displacement in metres from the centre of the bridge.-
(a) Interpret the meaning of the constant term $200$ in the model.
(b) Use the model to find the two values of $x$ at which the height is $346 \mathrm{~m}$.
(c) Given that the towers at each end are $346 \mathrm{~m}$ tall, use your answer to part $\mathbf{b}$ to calculate the length of the bridge to the nearest metre.
Solution$\textbf{(a) }\quad$ The bridge is $200 \mathrm{~m}$ above water level, since this is the height at the centre of the bridge.$\begin{aligned} &\\ \textbf{(b) }\quad 0.00012 x^{2}+200&=346\\\\ 12 x^{2}+200 &=346 \\\\ 0.00012 x^{2} &=146 \\\\ x^{2} &=\frac{146}{0.00012} \\\\ x &=\pm \sqrt{\frac{146}{0.00012}}\\\\ \therefore\ x=1103 \text{ or }x&=-1103\\\\ \end{aligned}$ So the height of the bridge is $346 \mathrm{~m}$, at the point $1103 \mathrm{~m}$ from the centre of the bridge. $\begin{aligned} &\\ \textbf{(c) }\quad \textbf{length of tower }&=2 \times 1103\\\\ &=2206 \mathrm{~m} \end{aligned}$ |
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Question 2
A diver launches herself off a springboard. The height of the diver, in metres, above the pool $t$ seconds after launch can be modelled by the following function: $$ h(t)=5 t-10 t^{2}+10, t \geq 0 $$(b) Use the model to find the time at which the diver hits the water.
(c) Rearrange $h(t)$ into the form $A-B(t-C)^{2}$ and give the values of the constants $A, B$ and $C$.
(d) Using your answer to part c or otherwise, find the maximum height of the diver, and the time at which this maximum height is reached.
Solution$\begin{aligned} \text{(a) }\quad &\text{height of springboard above water}\\\\ &=h(0) \\\\ &=10 \mathrm{~m} \\\\F \end{aligned}$$\begin{aligned} \text{(b) }\quad \text{When the diver } & \text{ hits water,}\\\\ h(t)&=0 \\\\ 5 t-10 t^{2}+10&=0 \\\\ t^{2}-\frac{t}{2}-1&=0 \\\\ t^{2}-\frac{t}{2}&=7\\\\ \left(t^{2}-\frac{t}{2}+\frac{1}{16}\right) &=1+\frac{1}{16} \\\\ \left(t-\frac{1}{4}\right)^{2} &=\frac{17}{16} \\\\ \therefore\ t &=\frac{1+\sqrt{17}}{4} \quad(\because t \geq 0) \\\\ &=1.3 \mathrm{~s}\\\\ \end{aligned}$ $\begin{aligned} h(t) &=5 t-10 t^{2}+10 \\\\ &=-10\left(t^{2}-\frac{1}{2} t\right)+10 \\\\ &=-10\left(t^{2}-\frac{1}{2} t+\frac{1}{16}\right)+10+\frac{10}{16} \\\\ &=-10\left(t-\frac{1}{4}\right)^{2}+\frac{85}{8} \\\\ &=\frac{85}{8}-10\left(t-\frac{1}{4}\right)^{2} \\\\ \end{aligned}$ $\therefore A -B(t-C)^{2}=\dfrac{85}{8}-10\left(t-\frac{1}{4}\right)^{2}$ $\begin{aligned} &\\ \therefore A &=\frac{85}{8}=10.625 \\\\ B &=10 \\\\ C &=\frac{1}{4}=0.25\\\\ \end{aligned}$ The vertex of graph is at $(C, A)=(0.25,10.625)\\\\ $. $\therefore$ At $t=0.25 \mathrm{~s}$, the maximum height of the diver is $10.625 \mathrm{~m}$ |
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Question 3
A car manufacturer uses a model to predict the fuel consumption, $y$ miles per gallon $(\mathrm{mpg})$, for a specific model of car travelling at a speed of $x \mathrm{~mph}$. $$ y=-0.01 x^{2}+0.975 x+16, x>0 $$-
(a) Use the model to find two speeds at which the car has a fuel consumption of $32.5 \mathrm{~mpg}$.
(b) Rewrite $y$ in the form $A-B(x-C)^{2}$, where $A, B$ and $C$ are constants to be found.
(c) Using your answer to part $b$, find the speed at which the car has the greatest fuel efficiency.
(d) Use the model to calculate the fuel consumption of a car travelling at $120 \mathrm{~mph}$. Comment on the validity of using this model for very high speeds.
Solution$\begin{aligned} \text {(a) } \hspace{1.5cm} y &=-0.01 x^{2}+0.975 x+16 \\\\ y &=\text { fuel consumption in (mpg)} \\\\ x &=\text { speed (mph) } \\\\ \text { When } y &=32.5 \\\\ 32.5 &=-0.01 x^{2}+0.975 x+16 \\\\ 0 &=-0.01 x^{2}+0.975 x-16.5 \\\\ 0 &=x^{2}-97.5 x+1650\\\\ \therefore x&=21.8 \text{ or } x=75.7\\\\ \end{aligned}$$\therefore$ The car has fuel consumption $32.5 \mathrm{mpg}$ at the speed $21.8$ mph or $75.7$ mph. $\begin{aligned} &\\ \text {(b) }\quad y &=-0.01\left(x^{2}-97.5 x\right)+16 \\\\ &=-0.01\left(x^{2}-97.5 x+48.75^{2}\right)+16+0.01(48.75)^{2} \\\\ &=-0.01(x-48.75)^{2}+39.77 \\\\ &=39.77-0.01(x-48.75)^{2} \\\\ \therefore A &-B(x-C)^{2}=39.77-0.01(x-48.75)^{2} \\\\ \therefore A &=39.77, B=0.01, C=48.75\\\\ \end{aligned}$ $\text {(c) }\quad $ The speed at which the car has greatest fuel efficiency is $48.75$ mph. $\begin{aligned} &\\ \text {(d) }\quad \text{ When } x&=120 \mathrm{mph},\\\\ y &=39.77-0.01(120-48.75)^{2} \\\\ &=-10.99 \\\\ &=-11 \text { mpg }\\\\ \end{aligned}$ Since the negative fuel consumption is impossible, this model is not valid for very high speeds. |
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Question 4
A fertiliser company uses a model to determine how the amount of fertiliser used, $f$ kilograms per hectare, affects the grain yield $g$, measured in tonnes per hectare. $$ g=6+0.03 f-0.00006 f^{2} $$-
(a) According to the model, how much grain would each hectare yield without any fertiliser?
(b) One farmer currently uses $20$ kilograms of fertiliser per hectare. How much more fertiliser would he need to use to increase his grain yield by 1 tonne per hectare?
Solution$\text{(a) } \quad g=$ the grain yield in ton/hectare.$\qquad f=$ amount of fertiliser used in $\mathrm{kg} /$ hectare $\begin{aligned} &\\ \text{(b) }\ g&=6+0.03f-0.00006 f^{2}\\\\ &\text{grain yield withont fertilizer}\\\\ &=6+0.03(0)-0.00006(0)^{2} \\\\ &=6 \text { ton/hactare}\\\\ \quad\text{When } f&=20,\\\\ g &=6+0.03(20)-0.00006(20)^{2} \\\\ &=6.576\\\\ \end{aligned}$ $\text{(c) }\ $ required amount of grain yield $=6.576+1$ tonnes $\begin{aligned} &\\ 6+0.03 f-0.00006 f^{2}&=7.576\\\\ \therefore\ 0.00006 f^{2}-0.03 f+1.576&=0\\\\ f=440.4 \text { (or) (reject) } f&=59.6 \\\\ \end{aligned}$ $\begin{aligned} \therefore\ &\text{ amount of extra fertiliser needed}\\\\ &=59.6-20 \\\\ &=39.6 \mathrm{~kg} / \text { hectare } \end{aligned}$ |
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Question 5
A spear is thrown over level ground from the top of a tower. The height, in metres, of the spear above the ground after $t$ seconds is modelled by the function: $$ h(t)=12.25+14.7 t-4.9 t^{2}, t \geq 0 $$-
(a) Interpret the meaning of the constant term $12.25$ in the model.
(b) After how many seconds does the spear hit the ground?
(c) Write $\mathrm{h}(t)$ in the form $A-B(t-C)^{2}$, where $A, B$ and $C$ are constants to be found. (d) Using your answer to part (c) or otherwise, find the maximum height of the spear above the ground, and the time at which this maximum height is reached.
Solution$\begin{aligned} \text{(a) }\quad &h(t)=12.25+14.7 t-4.9 t^{2}, t \geq 0 \\\\ &t=0, h(t)=12.25 \mathrm{~m}\\\\ \end{aligned}$$\therefore$ The constant term $12.25 \mathrm{~m}$ is the height of tower. $\begin{aligned} &\\ \text{(b) }\quad &\text{When the spear hit the ground,}\\\\ &h(t)=0 . \\\\ &12.25+14.7 t-4.9 t^{2}=0 \\\\ &t=-0.68 \text { or } t=3.68\\\\ \end{aligned}$ Since $t \geq 0$, the spear took $3.68 \mathrm{~s}$ to hit the ground. $\begin{aligned} &\\ \text{(c) }\quad h(t) &=12.25+14.7 t-4.9 t^{2} \\\\ &=-4.9\left(t^{2}-3 t\right)+12.25 \\\\ &=-4.9\left(t^{2}-3 t+1.5^{2}\right)+12.25+4.9\left(1.5^{2}\right) \\\\ &=-4.9(t-1.5)^{2}+23.275 \\\\ &=23.275-4.9(t-1.5)^{2} \\\\ \therefore\ A &-B(t-C)^{2}=23.275-4.9(t-1.5)^{2} \\\\ \therefore\ A &=23.275, B=4.9, C=1.5\\\\ \end{aligned}$ $\text{(d) }\quad$ The vertex of the gragh is $(C, A)=(1.5,23.275)\\\\ $ $\therefore$ When $t=1.58$, the spear reached the maximun height of $23.275 \mathrm{~m}$. |
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Question 6
For this question, $f(x)=4 k x^{2}+(4 k+2) x+1$, where $k$ is a real constant.-
(a) Find the discriminant of $f(x)$ in terms of $k$
(b) By simplifying your answer to part a or otherwise, prove that $\mathrm{f}(x)$ has two distinct real roots for all non-zero values of $k$.
(c) Explain why $f(x)$ cannot have two distinct real roots when $k=0$.
Solution$\begin{aligned} \text{(a) }\quad f(x)=4 k x^{2}+&(4 k+2) x+1 \\\\ \text { discriminant } &=(4 k+2)^{2}-4(4 k)(1) \\\\ &=16 k^{2}+16 k+4-16 k \\\\ &=4\left(4 k^{2}+1\right), k \neq 0\\\\ \end{aligned}$$\text{(b) }\quad$ Since $k^{2} \geq 0$ for all real values of $k$, $4\left(4 k^{2}+1\right)>0$ for all $x \in \mathbb{R}$. $\qquad\therefore(x)=0$ has always two distinct roots for all non-zero values of $k$. $\text{(c) }\quad$ When $k=0, f(x)=2 x+1$ which is a linear function with only one root. $\qquad\therefore f(x)$ can't have two distinct roots when $f(x)=0$. |
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စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!