The answer is A I just took the quiz
The y intercept would be when x=0,
so, y=0^2+0-6
this will make y= -6
y intercept is at (0,-6)
As for x intercept, this is quadratic so it will have two intercept where y=0
y=x^2+x-6
You'll have to break up the equation to find the answer
0=(x+3)(x-2)
x+3=0 leads to x= -3
x-2=0 leads to x= 2
so the x intercepts are at (2,0) and (-3,0)
The answer to putting them together-would be (2,5). And the equations would be y-y over x-x -6 and -11 being x and 5 and 3 being y
Answers:
1) (s o t)(x)=4x^2-x-4
2) (t o s)(x)=4(x-7)^2-(x-7)+3
Solution:
1) (s o t)(x)=s(t(x))=s(4x^2-x+3)=(4x^2-x+3)-7=4x^2-x+3-7
(s o t)(x)=4x^2-x-4
2) (t o s)(x)=t(s(x))=t(x-7)=4(x-7)^2-(x-7)+3
(t o s)(x)=4(x-7)^2-(x-7)+3
1.)
=(x-8i)(x+8i)
x^2+8ix-8ix-64i^2
x^2-64i^2
x^2-64(-1)
x^2+64
2.)
=(4x-7i)(4x+7i)
16x^2+28ix-28ix-49i^2
16x^2-49i^2
16x^2-49(-1)
16x^2+49
3.)
=(x+9i)(x+9i)
x^2+9ix+9ix+81i^2
x^2+18ix+81(-1)
x^2+18ix-81
4.)
=(x-2i)(x-2i)
x^2-2ix-2ix+4i^2
x^2-4ix+4(-1)
x^2-4ix-4
5.)
=[x+(3+5i)]^2
(x+5i+3)^2
(x+5i+3)(x+5i+3)
x^2+5ix+3x+5ix+25i^2+15i+3x+15i+9
x^2+6x+10ix+30i+25i^2+9
x^2+6x+10ix+30i+25(-1)+9
x^2+6x+10ix+30i-25+9
x^2+6x+10ix+30i-16
Hope this helps :)