http://prntscr.com/l6eh7i
@mhchen @kittybasil
mhchen is currently busy
\((5x^2-2x+1)-(3x-4)\) Distribute the - and then combine like terms
A?
Right
\(\large(\frac{b}{2})^2\) Knowing that \(f(x)=ax^2+bx+c\) Substitute and simplify
Im not sure can you explain pls
@dude
@mhchen
Sorry about that Im about to head out, do you know the b value of the equation they gave you?
No lolol
@mhchen can u help?
So like a perfect square trinomial is x + (2b)x + (b^2) Here 2b = 3. Find b^2
B?
or D?
none. Did you even try to solve for b in my equation?
If 2b = 3 b = (3/2) and b^2 = (3/2)^2 = (9/4) understand?
Ohhh okay I see, got it
okay so this one is a little bit harder. In order for it to have 2 real solutions. It has to be simplified into (x+a)(x+b) and to get there we need it in this form: ax^2 + bx + c = 0 X + Y = b X*Y = a*c
Here. b = -3. a = 1 If c = -4 then X could be -4 Y could be 1 X+Y = -3 X*Y = -4 so -4 would fit
A graph gets narrower when a gets higher: \[y= ax^2\] Which one of those is bigger than 5?
D
ye
Roots are where the line crosses the x-axis. Do you know where they are?
-2 and 3
yea
So the vertex of any quadratic function y = ax^2 + bx + c is \[\frac{-b}{2a}\] Here b = 4 a = -2 Can you find \[\frac{-b}{2a}?\]
1?
yes
So a side is 's' Another side is 's+16' Area = (s) * (s+16) which one of those is (s) * (s+16)?
im thinking D
Well actually s(s+16) = s*s + s*16 = s^2 + 16s
http://prntscr.com/l6fp63 http://prntscr.com/l6fpdu http://prntscr.com/l6fpg5 this is all one question
That last image. I assume it's d. Every time x increases by 4. y increases by 3. Also it's positive. And the y-intercept is 5.
@mhchen
The first one. if you look at the original graph. There's the points (0.-2) and (4.0) On the inverted graph it'll have (-2.0) and (0.4)
1 and 2?
there isnt a choice for 1 and 2
@Vocaloid
Oh I'm sorry I was thinking about whole numbers (my bad) It's only 2. natural numbers are 1 2 3 4 5... whole numbers are 0 1 2 3 4 5... I'm an idiot sorry
@dude @mhchen
Yeah they're right
Make a new post, this is getting too crowded
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