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I REALLY need your guys math help - ALG 2 - Quadratics
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Uh so basically I'm shitting the bed in ALG 2 and we got a take home quiz and I have no idea what the fuck I'm doing.
So I was given the function: f(x)= =3x^2-9x+2
(the ^2 is exponent)
and I need to find:
The y intercept
The axis of symmetry
The x-Coordinate of the vertex
Make a table of values that includes the vertex
Find the maximum and minimum values
and sketch the graph.
In all seriousness anyone who can answer all this correctly for me will get $5 in their paypal.
Post your reponses here
First come first serve.
Thanks NS
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y int = 2. just crunch the numbers man. that isn't to tough. or plug into a ti 89
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I tried but it gives me all this bull. Negatives and shit and doesn't seem right, I've got a D and I need this fucking grade I just want to make sure it's right.
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well to find the y intercept, you need to set the x's equal to 0. so the y intercept is 2.
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x coordinate of the vertex is -b/2a. then you plug that in to get the y value
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(Ill do it!!! here:
f(x)= =3x^2-9x+2
y-intercept = (0, 2)
The axis of symmetry
= x = 1.5
Vertex
= (1.5, -4.75)
Make a table of values that includes the vertex
x y
-1 14
0 2
1 -4
2 -4
1.5 -4.75
3 2
4 14
Find the maximum and minimum values
^(infinity = max and min?)
and sketch the graph.
(Use the table to graph it)
My paypal is kpazar@comcast.net
:D
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Alright im like 2 chapters ahead of you.
Minimum value - use your calculator
To find the vertex use this -
(-b)/(2a)
That will give you the x value of the vertex, and then you plug in that in the equation
Axis of symmetry is the x value you just found
And since its quadratic and since a is positive its going to have a minimum.
Im blanking on the y intercept though, sorry man
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table fix
x y
-1 14
0 2
1 -4
1.5 -4.75
2 -4
3 2
4 14
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critical points exist when the derivative of the function equals 0.
f(x)=3x^2-9x+2
f ' (x)= 9x-9
9x-9=0
x=1 is a critical point. now the problem should have an interval. evaluate the function at each point including the critical point. the highest is the max, lowest the min.
y intercept, using y=mx+b
find the slope using two points, for example (0,2) and (1,-4). slope= change of y/ change of x (5/6 in this case). now use the point (0,2) and m=5/6 to find b (the y intercept). 2= (5/6)*0 + b. b=2.
that is all for now
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so opposite of -9 is 9 divided by 2 times 3 gives you 9/6 as your x value for the vertex. then you plug it in, and you get 3(9/6)^2-9(9/6)+2. i dont have a calculator at hand, but thats your y value.
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i'm in algebra 2 but way too lazy to do that shit after finals
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just complete the square and the solves like most of it.
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Despite what some may think, skipping school actually improves my math skills..
:)
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leave feedback ahahahahaha
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