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those are my last two please........... ;/

# 4
D = b^2 - 4ac
D = (1)^2 - 4(0.05)(1)
D = ???

I'm sorry. I don't know any of this. I tried. I'm really really really sorry. :(((

im confused

y= 0.05 x^2 + 1x + 1 is your equation

I think phi has got you covered

hmmm....???

you mean that is how i do the quadratic part??

i have that part i need number 5

what did you get for a, b and c

wait what?? like for the explanation part??

D = (1)^2 - 4(0.05)(1)

??

a=0.05
b=1x
c=1

1^2-4(0.05)(1)

.8

now what is 1^2 ? 1*1 = 1
4*0.05 is 0.2
0.2*1 is 0.2
so
1- .2 = 0.8
the discriminant is 0.8

ok i get that thanks for actually working with me and not giving just the answer :D

so could you assist me in #5 as well??

a=0.05, b=1, c=1
\[x=\frac{−1±\sqrt{1^2- 4\times0.05\times1}}{2\times0.05}\]

yea i know it is a typo the graph is wrong

wait i am confused

which part?

the part like so i need to switch it to a -1?

Q1 general equation is y=0.05 x^2 -x +1
vertex form is 0.05(x-10)^2 - 4

so that means its this? y= 0.05 x^2 -1x + 1

in Q2, the -10 become +10
so that means its this? y= 0.05 x^2 -1x + 1 Yes

in Q2, the domain are all valid x's in this case, -inf to +infinity

in Q2, the range is all y values. from the graph, you see y is always bigger than -4

in Q3, your x-intercepts should be positive

in Q4, b is -1 not +1, but it won't change the discriminant calculation.

ok thanks for all that :D