+0

# Help ASAP

+5
103
3

Form quadratic functions having the following properties:

1.

2.

(These are two different questions)

Thank you

Guest Mar 4, 2017
Sort:

#1
0

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)

Guest Mar 4, 2017
#2
+4743
+1

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

hectictar  Mar 4, 2017
#3
0

(y=f(x))

Guest Mar 4, 2017

### 4 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details