Help ASAP

1. If zeroes are as above then

\(y=a(x-1+\sqrt{2})(x-1-\sqrt{2})\)

as \(y=a(x-b)(x-c)\)

where b and c are the zeroes

and as y-intercept = 4

\(4=a(-1+\sqrt{2})(-1-\sqrt{2}) \)

\( = a(-1)\)

Therefore a=-4, and thus

\(y=-4(x-1+\sqrt{2})(x-1-\sqrt{2})\)or

\(y=-4{x}^{2}-8x-4\)

(expanded)

2.As f(2)=0 and f(4)=0, 2 and 4 are zeroes, therefore

\(y=a(x-2)(x-4)\)

and as f(3)=-2, we can substitute x=3, y=-2 to solve for a

\(-2=a(1)(-1) \)therefore

a=2, and so

\(y=2(x-2)(x-4)\)or'

\(y=2{x}^{2}-12x+16\)

(expanded)

I honestly forgot how to to this so I had to refresh my memory with this website.

http://www.purplemath.com/modules/fromzero.htm

I'm sure it can explain how to do these better than I can. :)

1.

y = a[x - (1+√2)][x - (1 - √2)]

y = a[x - 1 - √2][x- 1 + √2]

y = a(x² - x + √2(x) - x + 1 - √2 - √2(x) + √2 - 2)

y = a(x² - 2x - 1)

Since the y-intercept is -4, x is 0 when y is -4. Use this information to find a.

-4 = a(0² - 2(0) - 1)

-4 = a(-1)

4 = a

So

y = 4(x² - 2x - 1)

y = 4x² - 8x - 4

Or you could say:

f(x) = 4x² - 8x - 4

-----------------------------------------------------

2.

The zeros are at x = 2 and x = 4.

y = a(x-2)(x-4)

y = a(x² - 6x + 8)

y is -2 when x is 3

-2 = a(3² - 6(3) + 8)

-2 = a(9 - 18 + 8)

-2 = a(-1)

2 = a

So

y = 2(x² - 6x + 8)

y = 2x² - 12x + 16

Or you could say:

f(x) = 2x² - 12x + 16

(y=f(x))