Question

one x-intercept for a parabola is at the point (3.22,0) find the other x-intercept for the parabola defined by this equation y=2x^2-8x+5 round to the nearest hundredth if necessary

Answer 1

(0.78,0)

Explanation:

I would use the quadratic formula.

`a=3`

`b=-8`

`c=5`

`\text{ The quadratic formula is } x=(-b \pm √(b^2-4ac))/(2a)\\\\\text{ Let's find } b^2-4ac \text{ first}\\(-8)^2-4(2)(5)\\64-8(5)\\64-40\\24\\\\\text{ Now let's find } -b\\-b=8\\\text{ And } 2a\\2(2)=4\\\\\text{ Let's plug in this information }\\x=(8 \pm √(24))/(4)\\\\\text{ We are now going to simplify }\\x=(8)/(4) \pm (√(24))/(4) \\x=2 \pm (√(4 \cdot 6))/(4)\\x=2 \pm (√(4) √(6))/(4) \\x=2 \pm (2 √(6))/(4)\\`

`x=2 \pm (√(6))/(2)`

So let's put both of these into out calculator

2 + sqrt(6)/2 and 2-sqrt(6)/2

One of them should be approximately 3.22 as the question suggests.

3.22 0.78

So the other x-intercept is approximately (0.78,0)