What are the x-intercepts of these parabolas?

Question

1. y= -x^2 +4x - 4

2. y= -x^2 +4x - 8

Thanks!

Answer 1

Hint: When the parabola intercepts the X axis, y is equal to zero, giving you an equation in the form 0=ax^2+bx+c. You can then factor or use the quadratic equation.  (-b +- sqrt(b^2-4ac)/2a

Simply by using part of the quadratic equation [sqrt of (b^2-4ac)] we can tell whether the parabola has 2, 1 or 0 intercepts.

1: a=-1, b=+4, c=-4

sqrt of (16 - (4 * -1 * -4)) = sqrt(16-16) = sqrt (0) = 0 so the parabola has only one intercept. The rest is easy, it's just -b/2a since the square root is zero, so the x intercept is at +2

2: a=-1, b=+4, c=-8

sqrt of (16 - (4 * -1 * -8)) = sqrt(16-32) = sqrt (-32) = imaginary so the parabola has no x intercept and we can stop here.

3: a=-1, b=+4, c=0

sqrt of (16 - (4 * -1 * 0)) = sqrt (16-0) = 4 so the parabola has two x intercepts. We now need to use the rest of the quadratic equation

(-4 + 4)/-2 = 0, (-4 -4)/-2 = 4, so the x intercepts are at 0 and 4

Answer 2

1. y= -x^2 +4x - 4

x-intercepts mean y=0

-x^2 +4x - 4=0

x^2 -4x + 4 = 0

(x-2)^2 = 0

x = 2 that mean it not intercepts but its touch the x



2. y= -x^2 +4x - 8

0 = x^2 -4x + 8

this as imaginary root

that mean no intersection

thanx