#### Toxxic Kitty

##### New member

- Joined
- Sep 9, 2015

- Messages
- 7

^{2}-3x+2

a) Find your exact x-intercepts

b) State your y-intercepts

c) Find your exact vertex

d) Graph f(x) and show and name all of the above information

So far I have the following:

a) Imaginary:

f(x) = 2x

^{2}--3x+2 x= [-b +/- √b

^{2}-4ac]/2a a= 2 b= -3 c=2

x= [-(-3) +/-√-3

^{2}-4(2)(2)]/2(2) --> x= [3 +/- √9 - 16]/4 --> x= [3 +/- √-7]/4 --> x= [3 +/- √-1 * √7]/4

x= 3i + √7 /4 , 3i - √7/4

b) (0,2)

f(x)= 2x

^{2}-3x+2

f(0) = 2(0)

^{2}-3(0)+2 --> f(0) = 0-0+2

f(0) = 2

c) (3/4, 2)

f(x) = 2x

^{2}-3x+2 ax

^{2}+bx+c a= 2 b= -3 (x,y)= x=(-b/2a) y= f((-b/2a))

x= [-(-3)/2(2)] --> x= (3/4) y= f(3/4) = 2(3/4)

^{2}-3(3/4)+2 --> f(3/4)= 9/4-9/4+2 --> f(3/4) = 2

So, now I'm a the graphing part. I know that since my x intercepts are imaginary, that I should only have to plot the vertex. I remember my professor saying that with an equation like this the graph would have two parabolas facing in opposite direction, that do not touch the x axis. I'm trying to figure out how to get the values for the second parabola, and if the work I did for a - c is correct.