11.3: Parabola
- Page ID
- 176812
Mwishoni mwa sehemu hii, utaweza:
- Grafu parabolas wima
- Grafu ya usawa parabolas
- Tatua programu na parabolas
Kabla ya kuanza, fanya jaribio hili la utayari.
- Grafu:\(y=-3 x^{2}+12 x-12\).
Kama amekosa tatizo hili, mapitio Mfano 9.47. - Tatua kwa kukamilisha mraba:\(x^{2}-6 x+6=0\).
Kama amekosa tatizo hili, mapitio Mfano 9.12. - Andika kwa fomu ya kawaida:\(y=3 x^{2}-6 x+5\).
Kama amekosa tatizo hili, mapitio Mfano 9.59.
Grafu Parabolas ya Wima
Sehemu inayofuata ya conic tutaangalia ni parabola. Tunafafanua parabola kama pointi zote katika ndege ambazo ni umbali sawa kutoka kwa uhakika uliowekwa na mstari uliowekwa. Hatua iliyowekwa inaitwa lengo, na mstari uliowekwa huitwa directrix ya parabola.
Parabola ni pointi zote katika ndege ambayo ni umbali sawa kutoka kwa uhakika uliowekwa na mstari uliowekwa. Hatua iliyowekwa inaitwa lengo, na mstari uliowekwa huitwa directrix ya parabola.
Hapo awali, tulijifunza kupiga parabolas wima kutoka kwa fomu ya jumla au fomu ya kawaida kwa kutumia mali. Njia hizo pia zitafanya kazi hapa. Sisi muhtasari mali hapa.
Parabolas ya wima
Fomu ya jumla \(y=a x^{2}+b x+c\) |
Fomu ya Standard \(y=a(x-h)^{2}+k\) |
|
---|---|---|
Mwelekeo | \ (y=a x^ {2} +b x+c\) ">\(a>0\) up;\(a<0\) chini | \ (y=a (x-h) ^ {2} +k\) ">\(a>0\) up;\(a<0\) chini |
Axis ya Ulinganifu | \ (y=a x^ {2} +b x+c\) ">\(x=-\dfrac{b}{2 a}\) | \ (y=a (x-h) ^ {2} +k\) ">\(x=h\) |
Vertex | \ (y=a x^ {2} +b x+c\) ">Mbadala\(x=-\dfrac{b}{2 a}\) na kutatua\(y .\) |
\ (y=a (x-h) ^ {2} +k\) ">\((h, k)\) |
\(y\)-kukatiza | \ (y = x^ {2} +b x+c\) ">Hebu\(x=0\) | \ (y=a (x-h) ^ {2} +k\) ">Hebu\(x=0\) |
\(x\)-hukataa | \ (y = x^ {2} +b x+c\) ">Hebu\(y=0\) | \ (y=a (x-h) ^ {2} +k\) ">Hebu\(y=0\) |
Grafu zinaonyesha nini parabolas inaonekana kama wakati wa kufungua au chini. Msimamo wao kuhusiana na\(x\) - au\(y\) -axis ni mfano tu.
Ili kuchora parabola kutoka kwa fomu hizi, tulitumia hatua zifuatazo.
Jinsi ya Grafu Parabolas Vertical\(y=a x^{2}+b x+c\) au\(f(x)=a(x-h)^{2}+k\) kutumia Mali.
- Hatua ya 1: Kuamua kama parabola inafungua juu au chini.
- Hatua ya 2. Pata mhimili wa ulinganifu.
- Hatua ya 3. Pata vertex.
- Hatua ya 4. Pata\(y\) -intercept. Pata uhakika ulinganifu kwa\(y\) -intercept katika mhimili wa ulinganifu.
- Hatua ya 5. Kupata\(x\) -intercepts.
- Hatua ya 6. Grafu parabola.
Mfano unaofuata unaangalia njia ya kuchora parabola kutoka kwa fomu ya jumla ya equation yake.
Grafu\(y=-x^{2}+6 x-8\) kwa kutumia mali.
Suluhisho:
\( \begin{align*} \color{red}{y} &\color{red}{=} a x^{2}+b x+c \\[4pt] \color{black}{y} &=-x^{2}+6 x-8 \end{align*}\) | |
Kwa kuwa\(a\) ni\(-1\), parabola inafungua chini. | |
Ili kupata mhimili wa ulinganifu, tafuta\(x=-\dfrac{b}{2 a}\). | \( \begin{align*} x &=-\dfrac{b}{2 a}\\[4pt] x &=-\dfrac{6}{2(-1)} \\[4pt] x &= 3 \end{align*}\) |
Mhimili wa ulinganifu ni\(x=3\). | |
Vertex iko kwenye mstari\(x=3\). | \(y=-x^{2}+6 x-8\) |
Hebu\(x=3\). | |
\(\begin{align*} y &=-9+18-8 \\[4pt] y &=1 \end{align*}\) | |
Vertex ni\((3,1)\). | |
\(y\)-Intercept hutokea wakati\(x=0\). | \(y=-x^{2}+6 x-8\) |
Mbadala\(x=0\). | \(y=-\color{red}{0}^{\color{black}{2}}+6 \cdot \color{red}{0} \color{black}{-} 8\) |
Kurahisisha. | \(y=-8\) |
\(y\)Kizuizi ni\((0,-8)\). | |
Hatua\((0,−8)\) ni vitengo vitatu upande wa kushoto wa mstari wa ulinganifu. Hatua vitengo vitatu kwa haki ya mstari wa ulinganifu ni\((6,−8)\). | Point ulinganifu kwa\(y\) -intercept ni\((6,−8)\). |
\(x\)-Intercept hutokea wakati\(y=0\). | \(y=-x^{2}+6 x-8\) |
Hebu\(y=0\). | \(\color{red}{0} \color{black}{=}-x^{2}+6 x-8\) |
Sababu ya GCF. | \(0=-\left(x^{2}-6 x+8\right)\) |
Sababu ya trinomial. | \(0=-(x-4)(x-2)\) |
Kutatua kwa\(x\). | \(x=4, \quad x=2\) |
\(x\)-intercepts ni\((4,0),(2,0)\). | |
Grafu parabola. |
Grafu\(y=-x^{2}+5 x-6\) kwa kutumia mali.
- Jibu
Grafu\(y=-x^{2}+8 x-12\) kwa kutumia mali.
- Jibu
Mfano unaofuata unaangalia njia ya kuchora parabola kutoka kwa fomu ya kawaida ya equation yake,\(y=a(x-h)^{2}+k\).
Andika\(y=3 x^{2}-6 x+5\) kwa fomu ya kawaida na kisha utumie mali ya fomu ya kawaida ili graph equation.
Suluhisho:
Andika upya kazi kwa\(y=a(x-h)^{2}+k\) fomu kwa kukamilisha mraba. | \(\begin{align*} y &=3 x^{2}-6 x+5 \\[4pt] y &=3\left(x^{2}-2 x\right)+5 \\[4pt] y &=3\left(x^{2}-2 x+1\right) + 5-3 \\[4pt] y &=3(x-1)^{2}+2 \end{align*}\) |
Tambua mara kwa mara\(a, h, k\). | \(a=3, h=1, k=2\) |
Tangu\(a=2\), parabola inafungua juu. | |
Mhimili wa ulinganifu ni\(x=h\). | Mhimili wa ulinganifu ni\(x=1\). |
Vertex ni\((h,k)\). | Vertex ni\((1,2)\). |
Pata\(y\) -intercept kwa kubadilisha\(x=0\), | \( \begin{align*} y &=3(x-1)^{2}+2 \\[4pt] y &=3 \cdot 0^{2}-6 \cdot 0+5 \\[4pt] y &=0 \end{align*} \) |
\(y\)-kukatiza\((0,5)\) | |
Pata hatua ya\((0,5)\) ulinganifu kwenye mhimili wa ulinganifu. | \((2,5)\) |
Kupata\(x\) -intercepts. | \(\begin{aligned} y &=3(x-1)^{2}+2 \\[4pt] 0 &=3(x-1)^{2}+2 \\[4pt] -2 &=3(x-1)^{2} \\[4pt] -\dfrac{2}{3} &=(x-1)^{2} \\[4pt] \pm \sqrt{-\dfrac{2}{3}} &=x-1 \end{aligned}\) |
Mzizi wa mraba wa nambari hasi inatuambia ufumbuzi ni namba ngumu. Kwa hiyo hakuna\(x\) -intercepts. | |
Grafu parabola. |
- Andika\(y=2 x^{2}+4 x+5\) kwa fomu ya kawaida na
- kutumia mali ya fomu ya kawaida kwa grafu equation.
- Jibu
-
- \(y=2(x+1)^{2}+3\)
- Andika\(y=-2 x^{2}+8 x-7\) kwa fomu ya kawaida na
- kutumia mali ya fomu ya kawaida kwa grafu equation.
- Jibu
-
- \(y=-2(x-2)^{2}+1\)
Grafu Ulalo Parabolas
Kazi yetu hadi sasa imeshughulikiwa tu na parabolas inayofungua au chini. Sasa tutaangalia parabolas ya usawa. Parabolas hizi zinafungua ama kushoto au kulia. Kama sisi interchange\(x\) na\(y\) katika equations yetu ya awali kwa parabolas, sisi kupata equations kwa parabolas kwamba wazi kwa upande wa kushoto au kulia.
Parabolas ya usawa
Fomu ya jumla \(x=a y^{2}+b y+c\) |
Fomu ya kawaida \(x=a(y-k)^{2}+h\) |
|
---|---|---|
Mwelekeo | \ (x=a y^ {2} +b y+c\) ">\(a>0\) haki;\(a<0\) kushoto | \ (x=a (y-k) ^ {2} +h\) ">\(a>0\) haki;\(a<0\) kushoto |
Axis ya Ulinganifu | \ (x=a y ^ {2} +b y+c\) ">\(y=-\dfrac{b}{2 a}\) | \ (x=a (y-k) ^ {2} +h\) ">\(y=k\) |
Vertex | \ (x=a y ^ {2} +b y+c\) ">Mbadala\(y=-\dfrac{b}{2 a}\) na kutatua\(x .\) |
\ (x=a (y-k) ^ {2} +h\) ">\((h, k)\) |
\(x\)-hukataa | \ (x=a y ^ {2} +b y+c\) ">Hebu\(x=0\) | \ (x=a (y-k) ^ {2} +h\) ">Hebu\(x=0\) |
\(y\)-kukatiza | \ (x=a y ^ {2} +b y+c\) ">Hebu\(y=0\) | \ (x=a (y-k) ^ {2} +h\) ">Hebu\(y=0\) |
Grafu zinaonyesha nini parabolas inaonekana kama wakati wa kushoto au kulia. Msimamo wao kuhusiana na\(x\) - au\(y\) -axis ni mfano tu.
Kuangalia parabolas hizi, je, grafu zao zinawakilisha kazi? Kwa kuwa grafu zote mbili bila kushindwa wima line mtihani, wao si kuwakilisha kazi.
Kwa graph parabola kwamba kufungua kwa upande wa kushoto au kulia kimsingi ni sawa na kile tulichofanya kwa parabolas kwamba kufungua juu au chini, na mabadiliko ya\(x\) na\(y\) vigezo.
- Hatua ya 1: Tambua kama parabola inafungua upande wa kushoto au kulia.
- Hatua ya 2: Pata mhimili wa ulinganifu.
- Hatua ya 3: Pata vertex.
- Hatua ya 4: Pata\(x\) -intercept. Pata uhakika ulinganifu kwa\(x\) -intercept katika mhimili wa ulinganifu.
- Hatua ya 5: Kupata\(y\) -intercepts.
- Hatua ya 6: Graph parabola.
Grafu\(x=2 y^{2}\) kwa kutumia mali.
Suluhisho:
Tangu\(a=2\), parabola inafungua kwa haki. | |
Ili kupata mhimili wa ulinganifu, tafuta\(y=-\dfrac{b}{2 a}\) | \(y=-\dfrac{b}{2 a}\) |
\(y=-\dfrac{0}{2(2)}\) | |
\(y=0\) | |
Mhimili wa ulinganifu ni\(y=0\). | |
Vertex iko kwenye mstari\(y=0\). | \(x=2 y^{2}\) |
Hebu\(y=0\). | |
\(x=0\) | |
Vertex ni\((0,0)\). |
Tangu vertex ni\((0,0)\), wote\(x\) - na\(y\) -intercepts ni uhakika\((0,0)\). Ili grafu ya parabola tunahitaji pointi zaidi. Katika kesi hii ni rahisi kuchagua maadili ya\(y\).
Sisi pia njama pointi symmetric\((2,1)\) na\((8,2)\) katika\(y\) -axis, pointi\((2,−1),(8,−2)\).
Grafu parabola.
Grafu\(x=y^{2}\) kwa kutumia mali.
- Jibu
Grafu\(x=-y^{2}\) kwa kutumia mali.
- Jibu
Katika mfano unaofuata, vertex sio asili.
Grafu\(x=-y^{2}+2 y+8\) kwa kutumia mali.
Suluhisho:
Tangu\(a=-1\), parabola inafungua upande wa kushoto. | |
Ili kupata mhimili wa ulinganifu, tafuta\(y=-\dfrac{b}{2 a}\) |
\(y=-\dfrac{b}{2 a}\) |
\(y=-\dfrac{2}{2(-1)}\) | |
\(y=1\) | |
Mhimili wa ulinganifu ni\(y=1\). | |
Vertex iko kwenye mstari\(y=1\). | \(x=-y^{2}+2 y+8\) |
Hebu\(y=1\). | |
\(x=9\) | |
Vertex ni\((9,1)\). | |
\(x\)-Intercept hutokea wakati\(y=0\). | \(x=-y^{2}+2 y+8\) |
\(x=8\) | |
\(x\)Kizuizi ni\((8,0)\). | |
Hatua\((8,0)\) ni kitengo kimoja chini ya mstari wa ulinganifu. Hatua ya ulinganifu kitengo kimoja juu ya mstari wa ulinganifu ni\((8,2)\) |
Hatua ya ulinganifu ni\((8,2)\). |
\(y\)-Intercept hutokea wakati\(x=0\). | \(x=-y^{2}+2 y+8\) |
Mbadala\(x=0\). | \(0=-y^{2}+2 y+8\) |
Kutatua. | \(y^{2}-2 y-8=0\) |
\((y-4)(y+2)=0\) | |
\(y=4, \quad y=-2\) | |
\(y\)-intercepts ni\((0,4)\) na\((0,-2)\). | |
Unganisha pointi kwa graph parabola. |
Grafu\(x=-y^{2}-4 y+12\) kwa kutumia mali.
- Jibu
Grafu\(x=-y^{2}+2 y-3\) kwa kutumia mali.
- Jibu
Katika Jedwali 11.2.4, tunaona uhusiano kati ya equation katika fomu ya kawaida na mali ya parabola. Jinsi ya sanduku inaorodhesha hatua za kuchora parabola katika fomu ya kawaida\(x=a(y-k)^{2}+h\). Tutatumia utaratibu huu katika mfano unaofuata.
Grafu\(x=2(y-2)^{2}+1\) kutumia mali.
Suluhisho:
Tambua mara kwa mara\(a, h, k\). | \(a=2, h=1, k=2\) |
Tangu\(a=2\), parabola inafungua kwa haki. | |
Mhimili wa ulinganifu ni\(y=k\). | Mhimili wa ulinganifu ni\(y=2\). |
Vertex ni\((h,k)\). | Vertex ni\((1,2)\). |
Pata\(x\) -intercept kwa kubadilisha\(y=0\). | \(x=2(y-2)^{2}+1\) \(x=2(0-2)^{2}+1\) \(x=9\) |
\(x\)Kizuizi ni\((9,0)\). | |
Pata hatua ya\((9,0)\) ulinganifu kwenye mhimili wa ulinganifu. | \((9,4)\) |
Kupata\(y\) -intercepts. Hebu\(x=0\). | \(\begin{aligned} x &=2(y-2)^{2}+1 \\ 0 &=2(y-2)^{2}+1 \\-1 &=2(y-2)^{2} \end{aligned}\) |
Mraba hauwezi kuwa hasi, kwa hiyo hakuna suluhisho halisi. Kwa hiyo hakuna\(y\) -intercepts. | |
Grafu parabola. |
Grafu\(x=3(y-1)^{2}+2\) kutumia mali.
- Jibu
Grafu\(x=2(y-3)^{2}+2\) kutumia mali.
- Jibu
Katika mfano unaofuata, tunaona a ni hasi na hivyo parabola inafungua upande wa kushoto.
Grafu\(x=-4(y+1)^{2}+4\) kutumia mali.
Suluhisho:
Tambua mara kwa mara\(a, h, k\). | \(a=-4, h=4, k=-1\) |
Tangu\(a=-4\), parabola inafungua upande wa kushoto. | |
Mhimili wa ulinganifu ni\(y=k\). | Mhimili wa ulinganifu ni\(y=-1\). |
Vertex ni\((h,k)\). | Vertex ni\((4,-1)\). |
Pata\(x\) -intercept kwa kubadilisha\(y=0\). | \(x=-4(y+1)^{2}+4\) \(x=-4(0+1)^{2}+4\) \(x=0\) |
\(x\)Kizuizi ni\((0,0)\). | |
Pata hatua ya\((0,0)\) ulinganifu kwenye mhimili wa ulinganifu. | \((0,-2)\) |
Kupata\(y\) -intercepts. | \(x=-4(y+1)^{2}+4\) |
Hebu\(x=0\). | \(\begin{aligned} 0 &=-4(y+1)^{2}+4 \\-4 &=-4(y+1)^{2} \\ 1 &=(y+1)^{2} \\ y+1 &=\pm 1 \end{aligned}\) |
\(y=-1+1 \quad y=-1-1\) | |
\(y=0 \quad\quad y=-2\) | |
\(y\)-intercepts ni\((0,0)\) na\((0,-2)\). | |
Grafu parabola. |
Grafu\(x=-4(y+2)^{2}+4\) kutumia mali.
- Jibu
Grafu\(x=-2(y+3)^{2}+2\) kutumia mali.
- Jibu
Mfano unaofuata unahitaji kwamba sisi kwanza kuweka equation katika fomu ya kawaida na kisha kutumia mali.
Andika\(x=2 y^{2}+12 y+17\) kwa fomu ya kawaida na kisha utumie mali ya fomu ya kawaida ili graph equation.
Suluhisho:
\(x=2 y^{2}+12 y+17\) | |
Andika upya kazi kwa\(x=a(y-k)^{2}+h\) fomu kwa kukamilisha mraba. | \(x=2\left(y^{2}+6 y\right)+17\) |
\(x=2(y+3)^{2}-1\) | |
Tambua mara kwa mara\(a, h, k\). | \(a=2, h=-1, k=-3\) |
Tangu\(a=2\), parabola inafungua kwa haki. | |
Mhimili wa ulinganifu ni\(y=k\). | Mhimili wa ulinganifu ni\(y=-3\). |
Vertex ni\((h,k)\). | Vertex ni\((-1,-3)\). |
Pata\(x\) -intercept kwa kubadilisha\(y=0\). | \(x=2(y+3)^{2}-1\) \(x=2(0+3)^{2}-1\) \(x=17\) |
\(x\)Kizuizi ni\((17,0)\). | |
Pata hatua ya\((17,0)\) ulinganifu kwenye mhimili wa ulinganifu. | \((17,-6)\) |
Kupata\(y\) -intercepts. Hebu\(x=0\). |
\(\begin{aligned} x &=2(y+3)^{2}-1 \\ 0 &=2(y+3)^{2}-1 \\ 1 &=2(y+3)^{2} \\ \dfrac{1}{2} &=(y+3)^{2} \\ y+3 &=\pm \sqrt{\dfrac{1}{2}} \\ y &=-3 \pm \dfrac{\sqrt{2}}{2} \end{aligned}\) |
\(y=-3+\dfrac{\sqrt{2}}{2} \quad y=-3-\dfrac{\sqrt{2}}{2}\) | |
\(y \approx-2.3 \quad y \approx-3.7\) | |
\(y\)-intercepts ni\(\left(0,-3+\dfrac{\sqrt{2}}{2}\right),\left(0,-3-\dfrac{\sqrt{2}}{2}\right)\). | |
Grafu parabola. |
- Andika\(x=3 y^{2}+6 y+7\) kwa fomu ya kawaida na
- Tumia mali ya fomu ya kawaida ili graph equation.
- Jibu
-
- \(x=3(y+1)^{2}+4\)
- Andika\(x=-4 y^{2}-16 y-12\) kwa fomu ya kawaida na
- Tumia mali ya fomu ya kawaida ili graph equation.
- Jibu
-
- \(x=-4(y+2)^{2}+4\)
Kutatua Maombi na Parabolas
Miundo mingi ya usanifu huingiza parabolas. Sio kawaida kwa madaraja ya kujengwa kwa kutumia parabolas kama tutakavyoona katika mfano unaofuata.
Kupata equation ya upinde parabolic sumu katika msingi wa daraja inavyoonekana. Andika equation katika fomu ya kawaida.
Suluhisho:
Sisi kwanza kuanzisha mfumo wa kuratibu na kuteka parabola. Grafu itatupa habari tunayohitaji kuandika equation ya grafu katika fomu ya kawaida\(y=a(x-h)^{2}+k\).
Hebu upande wa kushoto wa daraja uwe asili ya gridi ya kuratibu wakati huo\((0,0)\). Kwa kuwa msingi ni\(20\) miguu pana hatua\((20,0)\) inawakilisha upande wa chini wa kulia. Daraja lina urefu wa futi 10 kwenye hatua ya juu. Sehemu ya juu ni vertex ya parabola hivyo\(y\) -kuratibu ya |
|
Tambua vertex,\((h,k)\). | \((h, k)=(10,10)\) |
\(h=10, \quad k=10\) | |
Weka maadili katika fomu ya kawaida. Thamani ya bado\(a\) haijulikani. Ili kupata thamani ya\(a\) kutumia moja ya pointi nyingine kwenye parabola. |
\(\begin{aligned} y &=a(x-h)^{2}+k \\ y &=a(x-10)^{2}+10 \\(x, y) &=(0,0) \end{aligned}\) |
Badilisha maadili ya hatua nyingine katika equation. | \(y=a(x-10)^{2}+10\) \(0=a(0-10)^{2}+10\) |
Kutatua kwa\(a\). | \(\begin{aligned} 0 &=a(0-10)^{2}+10 \\-10 &=a(-10)^{2} \\-10 &=100 a \\ \dfrac{-10}{100} &=a \\ a &=-\dfrac{1}{10} \end{aligned}\) |
\(y=a(x-10)^{2}+10\) | |
Badilisha thamani ya\(a\) ndani ya equation. | \(y=-\dfrac{1}{10}(x-10)^{2}+10\) |
Kupata equation ya upinde parabolic sumu katika msingi wa daraja inavyoonekana. Andika equation katika fomu ya kawaida.
- Jibu
-
\(y=-\dfrac{1}{20}(x-20)^{2}+20\)
Kupata equation ya upinde parabolic sumu katika msingi wa daraja inavyoonekana. Andika equation katika fomu ya kawaida.
- Jibu
-
\(y=-\dfrac{1}{5} x^{2}+2 x y=-\dfrac{1}{5}(x-5)^{2}+5\)
Kupata rasilimali hizi online kwa maelekezo ya ziada na mazoezi na kazi quadratic na parabolas.
- Kazi za Quadratic
- Utangulizi wa Conics na Graphing Parabolas Horizontal
Dhana muhimu
- Parabola: parabola ni pointi zote katika ndege ambayo ni umbali sawa kutoka hatua ya kudumu na mstari fasta. Hatua iliyowekwa inaitwa lengo, na mstari uliowekwa huitwa directrix ya parabola.
Parabolas ya wima
Fomu ya jumla \(y=a x^{2}+b x+c\) |
Fomu ya Standard \(y=a(x-h)^{2}+k\) |
|
---|---|---|
Mwelekeo | \ (y=a x^ {2} +b x+c\) ">\(a>0\) up;\(a<0\) chini | \ (y=a (x-h) ^ {2} +k\) ">\(a>0\) up;\(a<0\) chini |
Axis ya Ulinganifu | \ (y=a x^ {2} +b x+c\) ">\(x=-\dfrac{b}{2 a}\) | \ (y=a (x-h) ^ {2} +k\) ">\(x=h\) |
Vertex | \ (y=a x^ {2} +b x+c\) ">Mbadala\(x=-\dfrac{b}{2 a}\) na kutatua\(y .\) |
\ (y=a (x-h) ^ {2} +k\) ">\((h, k)\) |
\(y\)-kukatiza | \ (y = x^ {2} +b x+c\) ">Hebu\(x=0\) | \ (y=a (x-h) ^ {2} +k\) ">Hebu\(x=0\) |
\(x\)-hukataa | \ (y = x^ {2} +b x+c\) ">Hebu\(y=0\) | \ (y=a (x-h) ^ {2} +k\) ">Hebu\(y=0\) |
- Jinsi ya kuchora parabolas wima\(y=a x^{2}+b x+c\) au\(f(x)=a(x-h)^{2}+k)\) kutumia mali.
- Kuamua kama parabola inafungua juu au chini.
- Pata mhimili wa ulinganifu.
- Pata vertex.
- Pata\(y\) -intercept. Pata uhakika ulinganifu kwa\(y\) -intercept katika mhimili wa ulinganifu.
- Kupata\(x\) -intercepts.
- Grafu parabola.
Parabolas ya usawa
Fomu ya jumla \(x=a y^{2}+b y+c\) |
Fomu ya kawaida \(x=a(y-k)^{2}+h\) |
|
---|---|---|
Mwelekeo | \ (x=a y^ {2} +b y+c\) ">\(a>0\) haki;\(a<0\) kushoto | \ (x=a (y-k) ^ {2} +h\) ">\(a>0\) haki;\(a<0\) kushoto |
Axis ya Ulinganifu | \ (x=a y ^ {2} +b y+c\) ">\(y=-\dfrac{b}{2 a}\) | \ (x=a (y-k) ^ {2} +h\) ">\(y=k\) |
Vertex | \ (x=a y ^ {2} +b y+c\) ">Mbadala\(y=-\dfrac{b}{2 a}\) na kutatua\(x .\) |
\ (x=a (y-k) ^ {2} +h\) ">\((h, k)\) |
\(x\)-hukataa | \ (x=a y ^ {2} +b y+c\) ">Hebu\(x=0\) | \ (x=a (y-k) ^ {2} +h\) ">Hebu\(x=0\) |
\(y\)-kukatiza | \ (x=a y ^ {2} +b y+c\) ">Hebu\(y=0\) | \ (x=a (y-k) ^ {2} +h\) ">Hebu\(y=0\) |
Jinsi ya grafu parabolas usawa\(x=a y^{2}+b y+c\) au\(x=a(y-k)^{2}+h\) kutumia mali.
- Tambua kama parabola inafungua upande wa kushoto au kulia.
- Pata mhimili wa ulinganifu.
- Pata vertex.
- Pata\(x\) -intercept. Pata uhakika ulinganifu kwa\(x\) -intercept katika mhimili wa ulinganifu.
- Kupata\(y\) -intercepts.
- Grafu parabola.
faharasa
- parabola
- Parabola ni pointi zote katika ndege ambayo ni umbali sawa kutoka kwa uhakika uliowekwa na mstari uliowekwa.