Question

Match the equations of parabolas with the x-intercepts of the parabolas. y = x2 + x − 12 y = x2 + 5x + 4 y = -2x2 + 11x + 8 y = -2x2 + 9x + 18 y = x2 − 5x − 24 y = -x2 + 5x + 14 (-2, 0), (7, 0) arrowRight (-4, 0), (3, 0) arrowRight (-4, 0), (-1, 0) arrowRight (-3, 0), (8, 0) arrowRight

Answer 1

Step-by-step explanation:

Match the equations of parabolas with the x-intercepts of the parabolas.

We find the x-intercept of the equations to match the points by putting y=0,

1) $y=x^2+x-12$

$x^2+x-12=0$

$x^2+4x-3x-12=0$

$x(x+4)-3(x+4)=0$

$(x+4)(x-3)=0$

$x=-4,3$

The x-intercept points are (-4,0) and (3,0).

2) $y=x^2+5x+4$

$x^2+5x+4=0$

$x^2+4x+x+4=0$

$x(x+4)+1(x+4)=0$

$(x+4)(x+1)=0$

$x=-4,-1$

The x-intercept points are (-4,0) and (-1,0).

3) $y=-2x^2+11x+8$

$-2x^2+11x+8=0$

$(x+0.65)(x-6.15)=0$

$x=-0.65,6.15$

The x-intercept points are (-0.65,0) and (6.15,0).

4) $y=-2x^2+9x+18$

$-2x^2+9x+18=0$

$(x+1.5)(x-6)=0$

$x=-1.5,6$

The x-intercept points are (-1.5,0) and (6,0).

5) $y=x^2-5x-24$

$x^2-5x-24=0$

$x^2-8x+3x-24=0$

$x(x-8)+3(x-8)=0$

$(x-8)(x+3)=0$

$x=8,-3$

The x-intercept points are (8,0) and (-3,0).

6) $y=-x^2+5x+14$

$-x^2+5x+14=0$

$-x^2+7x-2x+14=0$

$x(-x+7)+2(-x+7)=0$

$(x+2)(-x+7)=0$

$x=-2,7$

The x-intercept points are (-2,0) and (7,0).