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

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).