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?

GoldenLeaf Sep 3, 2015

#4**+5 **

Thankx Mr. Alan

I use the formula to make the graph and put the points on it.

Dragonlance Sep 4, 2015

#1**+5 **

The general expression for a quadratic function can be written as y = a*x^{2} + 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*0^{2} + b*0 + c or 1 = c

-2.5 = a*1^{2} + 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?

Alan Sep 3, 2015

#2**0 **

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?

-----------------------------------------

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

-----

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

Dragonlance Sep 3, 2015

#3**+5 **

You've gone astray by putting an explicit negative sign in for 'a'. Simpler to do the following:

.

Alan Sep 3, 2015

#4**+5 **

Best Answer

Thankx Mr. Alan

I use the formula to make the graph and put the points on it.

Dragonlance Sep 4, 2015