Question

Halla el eje de simetria y las coordenadas del vertice de cada una de las siguientes parabolas sin representarlas graficamente
a. y= 2x^2+4x-6
b. y= -x^2-8x+9
c. y= -2x^+x
d. y= x^+8x+15

Answer 1

Halla el eje de simetría y las coordenadas del vértice de cada una de las siguientes parábolas sin representarlas gráficamente
a. y= 2x^2+4x-6
$y = 0 2x^2+4x-6=0$
$\mathrm{Para\:}\quad a=2,\:b=4,\:c=-6:\quad x_{1,\:2}=\frac{-4\pm \sqrt{4^2-4\cdot \:2\left(-6\right)}}{2\cdot \:2}$
$x_{1}=\frac{-4+\sqrt{4^2-4\cdot \:2\left(-6\right)}}{2\cdot \:2}=\quad 1$
$x_{2}=\frac{-4-\sqrt{4^2-4\cdot \:2\left(-6\right)}}{2\cdot \:2}=\quad -3$
$\left(1,\:0\right),\:\left(-3,\:0\right)$
$x=0 \:y=2\cdot \:0^2+4\cdot \:0-6:\quad y=-6$
$\mathrm{X\:intersecta}:\:\left(1,\:0\right),\:\left(-3,\:0\right),\:\mathrm{Y\:intersecta}:\:\left(0,\:-6\right)$
$\mathrm{Para\:una\:parabola}\:ax^2+bx+c\:\mathrm{las\:x\:del\:vertice\:equivalen\:a}\:\frac{-b}{2a} a=2,\:b=4$
$x=\frac{-4}{2\cdot \:2} = -1 y=2\left(-1\right)^2+4\left(-1\right)-6 = -8$
$\mathrm{Por\:lo\:tanto,\:el\:vertice\:de\:la\:parabola\:es} \left(-1,\:-8\right)$

b. y= -x^2-8x+9
$y = 0 -x^2-8x+9=0$
$\mathrm{Para\:}\quad a=-1,\:b=-8,\:c=9:\quad x_{1,\:2}=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\left(-1\right)9}}{2\left(-1\right)}$
$x_{1}=\frac{-\left(-8\right)+\sqrt{\left(-8\right)^2-4\left(-1\right)9}}{2\left(-1\right)}:\quad -9$
$x_{2}=\frac{-\left(-8\right)-\sqrt{\left(-8\right)^2-4\left(-1\right)9}}{2\left(-1\right)}:\quad 1$
$\left(-9,\:0\right),\:\left(1,\:0\right)$
$x=0 y=-0^2-8\cdot \:0+9:\quad y=9$
$\mathrm{X\:intersecta}:\:\left(-9,\:0\right),\:\left(1,\:0\right),\:\mathrm{Y\:intersecta}:\:\left(0,\:9\right)$
$Para\:una\:parabola$
$a=-1,\:b=-8$
$x=\frac{-\left(-8\right)}{2\left(-1\right)} =-4$
$y=-\left(-4\right)^2-8\left(-4\right)+9=25 \mathrm{Por\:lo\:tanto,\:el\:vertice\:de\:la\:parabola\:es} \left(-4,\:25\right)$

c. y= -2x^2+x
$y=0 -2x^2+x=0$
$\mathrm{Para\:}\quad a=-2,\:b=1,\:c=0:\quad x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\left(-2\right)0}}{2\left(-2\right)}$
$x_{1}=\frac{-1+\sqrt{1^2-4\left(-2\right)0}}{2\left(-2\right)}:\quad 0$
$x_{2}=\frac{-1-\sqrt{1^2-4\left(-2\right)0}}{2\left(-2\right)}:\quad \frac{1}{2}$
$\left(0,\:0\right),\:\left(\frac{1}{2},\:0\right)$
$x=0 \:y=-2\cdot \:0^2+0:\quad y=0$
$\mathrm{X\:intersecta}:\:\left(0,\:0\right),\:\left(\frac{1}{2},\:0\right),\:\mathrm{Y\:intersecta}:\:\left(0,\:0\right)$
$Para\:una\:parabola$
$\:a=-2,\:b=1$
$x=\frac{-1}{2\left(-2\right)}=\frac{1}{4}$
$y=-2\left(\frac{1}{4}\right)^2+\frac{1}{4}=\frac{1}{8} \mathrm{Por\:lo\:tanto,\:el\:vertice\:de\:la\:parabola\:es} \left(\frac{1}{4},\:\frac{1}{8}\right)$

d. y= x^2+8x+15
$y=0 x^2+8x+15=0$
$\mathrm{Para\:}\quad a=1,\:b=8,\:c=15:\quad x_{1,\:2}=\frac{-8\pm \sqrt{8^2-4\cdot \:1\cdot \:15}}{2\cdot \:1}$
$x_{1}=\frac{-8+\sqrt{8^2-4\cdot \:1\cdot \:15}}{2\cdot \:1}:\quad -3$
$x_{2}=\frac{-8-\sqrt{8^2-4\cdot \:1\cdot \:15}}{2\cdot \:1}:\quad -5$
$\left(-3,\:0\right),\:\left(-5,\:0\right)$
$y=0 \:y=0^2+8\cdot \:0+15:\quad y=15$
$\mathrm{X\:intersecta}:\:\left(-3,\:0\right),\:\left(-5,\:0\right),\:\mathrm{Y\:intersecta}:\:\left(0,\:15\right)$
$\mathrm{Para\:una\:parabola} ,\:a=1,\:b=8$
$x=\frac{-8}{2\cdot \:1}=-4$
$y=\left(-4\right)^2+8\left(-4\right)+15=-1$
$\mathrm{Por\:lo\:tanto,\:el\:vertice\:de\:la\:parabola\:es} \left(-4,\:-1\right)$