Determine o valor de K para q o gráfico da funcão quadrática f(x)=(k+2)x(ao quadrado)+2x-k intercepte o eixo das abscissas em um único ponto
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Bom dia Izabel!
Solução!
![f(x)=(k+2) x^{2} +2x-k\\\\\\\
Fazendo!\\\\\\\<br /><br /><br />[tex]a=k+2\\\\\
b=2\\\\\\
c=-k\\\\\\\\\
\Delta=0\\\\\
b^{2}-4.a.c=0\\\\\
2^{2}-4(k+2).(-k)=0\\\\\\
4+(-4k-8).(-k)=0\\\\\\
4+4k^{2}+8k=0\\\\\\
4k^{2}+8k+4=0\\\\\\](https://tex.z-dn.net/?f=f%28x%29%3D%28k%2B2%29+x%5E%7B2%7D+%2B2x-k%5C%5C%5C%5C%5C%5C%5C%0AFazendo%21%5C%5C%5C%5C%5C%5C%5C%3Cbr+%2F%3E%3Cbr+%2F%3E%3Cbr+%2F%3E%5Btex%5Da%3Dk%2B2%5C%5C%5C%5C%5C%0Ab%3D2%5C%5C%5C%5C%5C%5C%0Ac%3D-k%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5CDelta%3D0%5C%5C%5C%5C%5C%0Ab%5E%7B2%7D-4.a.c%3D0%5C%5C%5C%5C%5C%0A2%5E%7B2%7D-4%28k%2B2%29.%28-k%29%3D0%5C%5C%5C%5C%5C%5C%0A4%2B%28-4k-8%29.%28-k%29%3D0%5C%5C%5C%5C%5C%5C%0A4%2B4k%5E%7B2%7D%2B8k%3D0%5C%5C%5C%5C%5C%5C%0A4k%5E%7B2%7D%2B8k%2B4%3D0%5C%5C%5C%5C%5C%5C+++%0A "f(x)=(k+2) x^{2} +2x-k\\\\\\\
Fazendo!\\\\\\\<br /><br /><br />[tex]a=k+2\\\\\
b=2\\\\\\
c=-k\\\\\\\\\
\Delta=0\\\\\
b^{2}-4.a.c=0\\\\\
2^{2}-4(k+2).(-k)=0\\\\\\
4+(-4k-8).(-k)=0\\\\\\
4+4k^{2}+8k=0\\\\\\
4k^{2}+8k+4=0\\\\\\")
![4k^{2}+8k+4=0\\\\\\
k= \dfrac{-b\pm \sqrt{b^{2}-4.a.c } }{2.a}\\\\\\\
k= \dfrac{-8\pm \sqrt{8^{2}-4.4.4 } }{2.4}\\\\\\\
k= \dfrac{-8\pm \sqrt{64-64 } }{8}\\\\\\\
k= \dfrac{-8\pm \sqrt{0} }{8}\\\\\\\
k= \dfrac{-8\pm 0}{8}\\\\\\\
k_{1}=k_{2}= \dfrac{-8}{8} =-1\\\\\\
\boxed{k_{1}=k_{2}=-1}](https://tex.z-dn.net/?f=4k%5E%7B2%7D%2B8k%2B4%3D0%5C%5C%5C%5C%5C%5C+%0Ak%3D+%5Cdfrac%7B-b%5Cpm+%5Csqrt%7Bb%5E%7B2%7D-4.a.c+%7D+%7D%7B2.a%7D%5C%5C%5C%5C%5C%5C%5C%0Ak%3D+%5Cdfrac%7B-8%5Cpm+%5Csqrt%7B8%5E%7B2%7D-4.4.4+%7D+%7D%7B2.4%7D%5C%5C%5C%5C%5C%5C%5C%0Ak%3D+%5Cdfrac%7B-8%5Cpm+%5Csqrt%7B64-64+%7D+%7D%7B8%7D%5C%5C%5C%5C%5C%5C%5C%0Ak%3D+%5Cdfrac%7B-8%5Cpm+%5Csqrt%7B0%7D+%7D%7B8%7D%5C%5C%5C%5C%5C%5C%5C%0Ak%3D+%5Cdfrac%7B-8%5Cpm+0%7D%7B8%7D%5C%5C%5C%5C%5C%5C%5C%0Ak_%7B1%7D%3Dk_%7B2%7D%3D+%5Cdfrac%7B-8%7D%7B8%7D+%3D-1%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7Bk_%7B1%7D%3Dk_%7B2%7D%3D-1%7D++%0A "4k^{2}+8k+4=0\\\\\\
k= \dfrac{-b\pm \sqrt{b^{2}-4.a.c } }{2.a}\\\\\\\
k= \dfrac{-8\pm \sqrt{8^{2}-4.4.4 } }{2.4}\\\\\\\
k= \dfrac{-8\pm \sqrt{64-64 } }{8}\\\\\\\
k= \dfrac{-8\pm \sqrt{0} }{8}\\\\\\\
k= \dfrac{-8\pm 0}{8}\\\\\\\
k_{1}=k_{2}= \dfrac{-8}{8} =-1\\\\\\
\boxed{k_{1}=k_{2}=-1}")
![Substituindo~~k~~na~~equac\~ao~~temos!\\\\\\
f(x)=(k+2) x^{2} +2x-k\\\\\\
f(x)=(-1+2) x^{2} +2x-(-1)\\\\\\
f(x)=x^{2} +2x+1\\\\\\\\
Justificativa: Sendo~~k=-1~~o~~vertice~~da~~parabola~~toca~~o~~eixo\\\\\\\
das~~abscissas~~no~~ponto~~-1.\\\\\\\
Bom~~ dia!\\\\\\
Bons~~ estudos!](https://tex.z-dn.net/?f=Substituindo%7E%7Ek%7E%7Ena%7E%7Eequac%5C%7Eao%7E%7Etemos%21%5C%5C%5C%5C%5C%5C%0Af%28x%29%3D%28k%2B2%29+x%5E%7B2%7D+%2B2x-k%5C%5C%5C%5C%5C%5C%0Af%28x%29%3D%28-1%2B2%29+x%5E%7B2%7D+%2B2x-%28-1%29%5C%5C%5C%5C%5C%5C%0Af%28x%29%3Dx%5E%7B2%7D+%2B2x%2B1%5C%5C%5C%5C%5C%5C%5C%5C%0AJustificativa%3A+Sendo%7E%7Ek%3D-1%7E%7Eo%7E%7Evertice%7E%7Eda%7E%7Eparabola%7E%7Etoca%7E%7Eo%7E%7Eeixo%5C%5C%5C%5C%5C%5C%5C%0Adas%7E%7Eabscissas%7E%7Eno%7E%7Eponto%7E%7E-1.%5C%5C%5C%5C%5C%5C%5C%0ABom%7E%7E+dia%21%5C%5C%5C%5C%5C%5C%0ABons%7E%7E+estudos%21 "Substituindo~~k~~na~~equac\~ao~~temos!\\\\\\
f(x)=(k+2) x^{2} +2x-k\\\\\\
f(x)=(-1+2) x^{2} +2x-(-1)\\\\\\
f(x)=x^{2} +2x+1\\\\\\\\
Justificativa: Sendo~~k=-1~~o~~vertice~~da~~parabola~~toca~~o~~eixo\\\\\\\
das~~abscissas~~no~~ponto~~-1.\\\\\\\
Bom~~ dia!\\\\\\
Bons~~ estudos!")
Solução!
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