So, I just tested into AP Calculus AB... and this problem comes up.
I have figured out the first one, but not the second. Anybody wanna help?
Thankx Mr. Alan
I use the formula to make the graph and put the points on it.
The general expression for a quadratic function can be written as y = a*x2 + b*x + c where a, b and c are constants.
You can find a, b and c by using the points on the graph
2 = a*(-2)2 + b*(-2) + c or 2 = 4a -2b + c
1 = a*02 + b*0 + c or 1 = c
-2.5 = a*12 + b*1 + c or -2.5 = a + b + c
You now have 3 equations for the 3 unknowns a, b and c. Can you take it from here?
Hi Mr. Alan
I been playing with this for 3 hours and this is what I have.
I guess you can read what I do here
But when I solve it and put in the x part it do not equal the y part.
Can you see what I do wrong?
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This is way ahead of where I am in maths. But it look like a fun one to do.
a quadratic is "y = ax2 + bx + c" a is negative because the parabola is upsidedown.
(-2,2)(0,1)(1,-2.5)
-a(-2)^2+b(-2)+c =2
-a(0)^2+b(0)+c =1
-a(1)^2+b(1)+c =-2.5
a(4)-2b+c=2
0+0+c=1
-a+b+c=-2.5
-----
a(4)-2b=1
a-b=1.5
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B=1.5-a
4a-2(1.5-a)=1
4a-3+a=1
5a=4
A=4/5
A=1.2
--------
B=1.5-1.2
B=0.3
--------
A=1.2 B=0.3 C=1
1.2(x^2)+0.3x+1=0
1.2(-2^2)+0.3(-2)+1 not = 0
You've gone astray by putting an explicit negative sign in for 'a'. Simpler to do the following:
.
Thankx Mr. Alan
I use the formula to make the graph and put the points on it.