Second derivative

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Loki123

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What am I supposed to do here?

The sum of the zeros of the function y = f (x) which contains the points A (2,6) and B (-1, -6) and which y '' = 2, is?

My thought was to use a formula for the slope but the formula for the slope with two points is the equation of a line and this might be a curve... Any advice? What should I look for?
 
Last edited:
Loki123 said:
What am I supposed to do here?

The sum of the zeros of the function y = f (x) which contains the points A (2,6) and B (-1, -6) and which y '' = 2, is?

My thought was to use a formula for the slope but the formula for the slope with two points is the equation of a line and this might be a curve... Any advice? What should I look for?
Click to expand...
If I were to do this problem, I would assume the function is y = x2 + B*x + C and solve.
 
Subhotosh Khan said:
If I were to do this problem, I would assume the function is y = x2 + B*x + C and solve.
Click to expand...
If i assume that
y'=2x+b

y=x^2+bx+c
(2,6)
6=4+2b+c
2=2b+c (1)

(-1,-6)
-6=1-b+c
-7=-b+c (2)

(2)-(1)
-9=-3b
b=3
2=6+c
c=-4

x^2+3x-4=0
(-3+/-sqrt(9+16))/2
(-3+/-5)/2
x1=1
X2=-4

X1+x2=-3 or -b/a=-3
What do you think?
 
Loki123 said:
If i assume that
y'=2x+b

y=x^2+bx+c
(2,6)
6=4+2b+c
2=2b+c (1)

(-1,-6)
-6=1-b+c
-7=-b+c (2)

(2)-(1)
-9=-3b
b=3
2=6+c
c=-4

x^2+3x-4=0
(-3+/-sqrt(9+16))/2
(-3+/-5)/2
x1=1
X2=-4

X1+x2=-3 or -b/a=-3
What do you think?
Click to expand...
You shouldn't have to assume that form. If you consider what I said in#2, you can show the function is in that form.
 
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