| ဖုန်းဖြင့်ကြည့်သည့်အခါ စာများကို အပြည့်မမြင်ရလျှင် screen ကို ဘယ်ညာ ဆွဲကြည့်နိုင်ပါသည်။ |
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| Quadratic Function (Standard Form) | $f(x)=a x^{2}+b x+c, a \neq 0$ |
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| Graph | Parabola $a>0$ (opens upward) $a<0$ (opens downward) |
| Axis of Symmetry | $x=-\displaystyle\frac{b}{2 a}$ |
| Vertex | $\left(-\displaystyle\frac{b}{2 a}, f\left(-\displaystyle\frac{b}{2 a}\right)\right)=\left(-\displaystyle\frac{b}{2 a},-\displaystyle\frac{b^{2}-4 a c}{4 a}\right)$ |
| y-intercept | (0, c) |
| Discriminant | $b^{2}-4 a c$ $b^{2}-4 a c>0 \Rightarrow$ two $x$ intercepts (cuts $x-$ axis at two points) $b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (touch $x$ -axis at one point $)$ $b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (does not intersect $x$ -axis) |
| Quadratic Equation | $a x^{2}+b x+c=0, a \neq 0$ |
| Quadratic Formula | $x=\displaystyle\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ |
| Quadratic Function (Vertex Form) | $f(x)=a(x-h)^{2}+k, a \neq 0$ |
| Vertex | $(h, k)$ |
| Axis of Symmetry (Vertex Form) | $x=h$ |
| Quadratic Function (Intercept Form when discriminant $>0$) | $f(x)=a(x-p)(x-q), a \neq 0$ |
| $X$ -intercept points | $(p, 0)$ and $(q, 0)$ |
| Axis of Symmetry | $x=\displaystyle\frac{p+q}{2}$ |
| Quadratic Inequality | $a x^{2}+b x+c>0$ $a x^{2}+b x+c \geq 0$ $a x^{2}+b x+c<0$ $a x^{2}+b x+c \leq 0$ |
$a>0$ and $b^2-4ac<0$ | The graph does not cut the $x$ -axis. $y<0 \Rightarrow$ solution set $=\varnothing$ $y=0 \Rightarrow$ solution set $=\varnothing$ $y>0 \Rightarrow$ solution set $=\mathbb{R}$ |
$a<0$ and $b^2-4ac<0$ | The graph does not cut the $x$ -axis. $y<0 \Rightarrow$ solution set $=\mathbb{R}$ $y=0 \Rightarrow$ solution set $=\varnothing$ $y>0 \Rightarrow$ solution set $=\varnothing$ |
$a>0$ and $b^2-4ac=0$ | The graph touches the $x$ -axis. $y<0 \Rightarrow$ solution set $=\varnothing$ $y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y>0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$ |
$a<0$ and $b^2-4ac=0$ | The graph touches the $x$ -axis. $y<0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y>0 \Rightarrow$ solution set $=\varnothing$ |
$a>0$ and $b^2-4ac>0$ | The graph cuts the $x$ -axis at two points. $y<0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$ $y=0 \Rightarrow$ solution set $=\{p, q\}$ $y>0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$ |
$a<0$ and $b^2-4ac>0$ | The graph cuts the $x$ -axis at two points. $y<0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$ $y=0 \Rightarrow$ solution set $=\{p, q\}$ $y>0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$ |
ယခု post တွင်ပါဝင်သော မေးခွန်းများသည် grade 10 သင်ရိုးသစ်ပြဌာန်းချက်နှင့် သက်ဆိုင်သော သင်ရိုးလေ့ကျင့်ခန်း ပြင်ပမေးခွန်းများ ဖြစ်ပြီး Target Mathematics Grade 10 (Volume 1) တွင် Miscellaneous Exercise အဖြစ်ထည့်သွင်းပေးခဲ့ပါသည်။
| (a) | What is the vertex of $y$? |
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| (b) | What are the $x$-intercepts of the graph of $y$ ? |
| (c) | Find the $x$-coordinates of the ponts on the graph $y = x^2 + 2x − 8$ when $y = −8$. |
| (a) | Find the amount, $x$, that the company has to spend to maximize its profit. |
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| (b) | Find the maximum profit. |
| ဖုန်းဖြင့်ကြည့်သည့်အခါ စာများကို အပြည့်မမြင်ရလျှင် screen ကို ဘယ်ညာ ဆွဲကြည့်နိုင်ပါသည်။ |
|---|
| Quadratic Function (Standard Form) | $f(x)=a x^{2}+b x+c, a \neq 0$ |
|---|---|
| Graph | Parabola $a>0$ (opens upward) $a<0$ (opens downward) |
| Axis of Symmetry | $x=-\displaystyle\frac{b}{2 a}$ |
| Vertex | $\left(-\displaystyle\frac{b}{2 a}, f\left(-\displaystyle\frac{b}{2 a}\right)\right)=\left(-\displaystyle\frac{b}{2 a},-\displaystyle\frac{b^{2}-4 a c}{4 a}\right)$ |
| y-intercept | (0, c) |
| Discriminant | $b^{2}-4 a c$ $b^{2}-4 a c>0 \Rightarrow$ two $x$ intercepts (cuts $x-$ axis at two points) $b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (touch $x$ -axis at one point $)$ $b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (does not intersect $x$ -axis) |
| Quadratic Equation | $a x^{2}+b x+c=0, a \neq 0$ |
| Quadratic Formula | $x=\displaystyle\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ |
| Quadratic Function (Vertex Form) | $f(x)=a(x-h)^{2}+k, a \neq 0$ |
| Vertex | $(h, k)$ |
| Axis of Symmetry (Vertex Form) | $x=h$ |
| Quadratic Function (Intercept Form when discriminant $>0$) | $f(x)=a(x-p)(x-q), a \neq 0$ |
| $X$ -intercept points | $(p, 0)$ and $(q, 0)$ |
| Axis of Symmetry | $x=\displaystyle\frac{p+q}{2}$ |
| Quadratic Inequality | $a x^{2}+b x+c>0$ $a x^{2}+b x+c \geq 0$ $a x^{2}+b x+c<0$ $a x^{2}+b x+c \leq 0$ |
| $a>0$ and $b^2-4ac<0$ | The graph does not cut the $x$ -axis. $y<0 \Rightarrow$ solution set $=\varnothing$ $y=0 \Rightarrow$ solution set $=\varnothing$ $y>0 \Rightarrow$ solution set $=\mathbb{R}$ |
| $a<0$ and $b^2-4ac<0$ | The graph does not cut the $x$ -axis. $y<0 \Rightarrow$ solution set $=\mathbb{R}$ $y=0 \Rightarrow$ solution set $=\varnothing$ $y>0 \Rightarrow$ solution set $=\varnothing$ |
| $a>0$ and $b^2-4ac=0$ | The graph touches the $x$ -axis. $y<0 \Rightarrow$ solution set $=\varnothing$ $y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y>0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$ |
| $a<0$ and $b^2-4ac=0$ | The graph touches the $x$ -axis. $y<0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y>0 \Rightarrow$ solution set $=\varnothing$ |
| $a>0$ and $b^2-4ac>0$ | The graph cuts the $x$ -axis at two points. $y<0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$ $y=0 \Rightarrow$ solution set $=\{p, q\}$ $y>0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$ |
| $a<0$ and $b^2-4ac>0$ | The graph cuts the $x$ -axis at two points. $y<0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$ $y=0 \Rightarrow$ solution set $=\{p, q\}$ $y>0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$ |
အထက်ဖော်ပြပါ quadratic function နှင့်ဆိုင်သော definitions နှင့် concepts များကိုသိရှိနားလည်ပြီးလျှင် အောက်ပါ MCQ များကို လေ့ကျင့် ဖြေဆိုနိုင်ပါပြီ။ ဖြေဆိုပြီးကြောင်း Submit လုပ်ပြီးလျှင် ရမှတ်နှင့် အဖြေမှန်ကိုပါ ပြပေးမည် ဖြစ်သည်။ ရှင်းလင်းချက်မပါဝင်ပါ။
| Quadratic Function (Standard Form) | $f(x)=a x^{2}+b x+c, a \neq 0$ |
|---|---|
| Graph | Parabola $a>0$ (opens upward) $a<0$ (opens downward) |
| Axis of Symmetry | $x=-\displaystyle\frac{b}{2 a}$ |
| Vertex | $\left(-\displaystyle\frac{b}{2 a}, f\left(-\displaystyle\frac{b}{2 a}\right)\right)=\left(-\displaystyle\frac{b}{2 a},-\displaystyle\frac{b^{2}-4 a c}{4 a}\right)$ |
| y-intercept | (0, c) |
| Discriminant | $b^{2}-4 a c$ $b^{2}-4 a c>0 \Rightarrow$ two $x$ intercepts (cuts $x-$ axis at two points) $b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (touch $x$ -axis at one point $)$ $b^{2}-4 a c=0 \Rightarrow$ one $x$ intercepts (does not intersect $x$ -axis) |
| Quadratic Equation | $a x^{2}+b x+c=0, a \neq 0$ |
| Quadratic Formula | $x=\displaystyle\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ |
| Quadratic Function (Vertex Form) | $f(x)=a(x-h)^{2}+k, a \neq 0$ |
| Vertex | $(h, k)$ |
| Axis of Symmetry (Vertex Form) | $x=h$ |
| Quadratic Function (Intercept Form when discriminant $>0$) | $f(x)=a(x-p)(x-q), a \neq 0$ |
| $X$ -intercept points | $(p, 0)$ and $(q, 0)$ |
| Axis of Symmetry | $x=\displaystyle\frac{p+q}{2}$ |
| Quadratic Inequality | $a x^{2}+b x+c>0$ $a x^{2}+b x+c \geq 0$ $a x^{2}+b x+c<0$ $a x^{2}+b x+c \leq 0$ |
| $a>0$ and $b^2-4ac<0$ | The graph does not cut the $x$ -axis. $y<0 \Rightarrow$ solution set $=\varnothing$ $y=0 \Rightarrow$ solution set $=\varnothing$ $y>0 \Rightarrow$ solution set $=\mathbb{R}$ |
| $a<0$ and $b^2-4ac<0$ | The graph does not cut the $x$ -axis. $y<0 \Rightarrow$ solution set $=\mathbb{R}$ $y=0 \Rightarrow$ solution set $=\varnothing$ $y>0 \Rightarrow$ solution set $=\varnothing$ |
| $a>0$ and $b^2-4ac=0$ | The graph touches the $x$ -axis. $y<0 \Rightarrow$ solution set $=\varnothing$ $y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y>0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$ |
| $a<0$ and $b^2-4ac=0$ | The graph touches the $x$ -axis. $y<0 \Rightarrow$ solution set $=\mathbb{R} \backslash\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y=0 \Rightarrow$ solution set $=\left\{-\displaystyle\frac{b}{2 a}\right\}$ $y>0 \Rightarrow$ solution set $=\varnothing$ |
| $a>0$ and $b^2-4ac>0$ | The graph cuts the $x$ -axis at two points. $y<0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$ $y=0 \Rightarrow$ solution set $=\{p, q\}$ $y>0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$ |
| $a<0$ and $b^2-4ac>0$ | The graph cuts the $x$ -axis at two points. $y<0 \Rightarrow$ solution set $=\{x \mid x<p$ or $x>q\}$ $y=0 \Rightarrow$ solution set $=\{p, q\}$ $y>0 \Rightarrow$ solution set $=\{x \mid p<x<q\}$ |
| Show that the infinite square root $\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}=\displaystyle\frac{1+\sqrt{5}}{2} .$ Solution Let $x=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}$ Squaring both sides, $x^{2}=1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}$ $\therefore x^{2}=1+x$ $\therefore x^{2}-x-1=0$ $\therefore x=\displaystyle\frac{1 \pm \sqrt{5}}{2}$ Since $x>0, x =\displaystyle\frac{1+\sqrt{5}}{2}$ $\therefore \quad \sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\ldots}}}}}=\displaystyle\frac{1+\sqrt{5}}{2} .$ |
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| 1. | Use completing the square to prove that $3 n^{2}-4 n+10$ is positive for all values of $n$. | |
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| 2. | Use completing the square to prove that $-n^{2}-2 n-3$ is negative for all values of $n$. | |
| 3. | Find the values of $k$ for which $x^{2}+6 x+k=0$ has two real solutions. | |
| 4. | Find the value of $t$ for which $2 x^{2}-3 x+t=0$ has exactly one solution. | |
| 5. | Given that the function $f(x)=s x^{2}+8 x+s$ has equal roots, find the value of the positive constant $s$. | |
| 6. | Find the range of values of $k$ for which $3 x^{2}-4 x+k=0$ has no real solutions. | |
| 7. | The function $g(x)=x^{2}+3 p x+(14 p-3)$, where $p$ is an integer, has two equal roots. | |
| (a) | Find the value of $p$. | |
| (b) | For this value of $p$, solve the equation $x^{2}+3 p x+(14 p-3)=0$. | |
| 8. | $h(x)=2 x^{2}+(k+4) x+k$, where $k$ is a real constant. | |
| (a) | Find the discriminant of $h(x)$ in terms of $k$. | |
| (b) | Hence or otherwise, prove that $h(x)$ has two distinct real roots for all values of $k$. | |
| 9. | The equation $p x^{2}-5 x-6=0$, where $p$ is a constant, has two distinct real roots. Prove that $p$ satisfies the inequality $p>-\displaystyle\frac{25}{24} $. | |
| $\begin{array}{ll} 1. & \text{Hint:}\ 3\left(n-\displaystyle\frac{2}{3}\right)^{2}+\displaystyle\frac{26}{3}\\\\ 2. & \text{Hint:}\ -(n+1)^2-2\\\\ 3. & k<9 \\\\ 4. & t=\displaystyle\frac{9}{8}\\\\ 5. & s=4\\\\ 6. & \left\{x\ |\ k>\displaystyle\frac{4}{3}\right\}\\\\ 7. & \text{(a)}\ p=6, \text{(b)}\ x=-9\\\\ 8. & \begin{array}{ll} \text{(a)}& \text{discriminant}=(k+4)^2-8k \\ \text{(b)}& \text{Hint} : (k+4)^2-8k=k^2+16>0\ \text{for all values of}\ k \end{array}\\\\ 9. & \text{Hint: discriminant} >0 \end{array}$ |
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