Question

The graph of a quadratic function is given. Determine the function's equation.

Answer 1

you have the three points of intercepts (0,1),(-1,0),(1,0)
just sub in the equation ax^2+bx+c=0 to see what the constants are!

Answer 2

Have you learned about factoring a quadratic equation ?

Answer 3

No

Answer 4

Often you are given an equation of the form
ax^2 +bx +c =0
and are asked to factor it into
(x- c)(x-d) =0
and then you do
x-c =0 --> which means x=c
and
x-d = 0 which means x= d

the point of all this is that c and d are the x values where the curve crosses the x-axis.

Answer 5

If you look at your graph, you can see where the graph crosses the x-axis

can you find those 2 x values ?

Answer 6

-2 and 2?

Answer 7

err -1 and 1 rather

Answer 8

yes, so x=-1 and x=+1 are the two spots where the curve gives y=0

x= -1 can be re-written as x+1 =0 (add +1 to both sides)
x= +1 is also x-1 = 0 (add -1 to both sides)
if we multiply (x+1)(x-1) we can say it still = 0
(x+1)(x-1)=0

finally, we can multiply both sides by an unknown number , call it "b"
b(x+1)(x-1) = 0*b = 0

the equation of your curve will be
y = b(x+1)(x-1)

can you expand (x+1)(x-1) ?

Answer 9

Uhhh I think so

Answer 10

x^2-1

Answer 11

so now we have
y= b(x^2-1)

you need to find "b"

one way is notice the curve goes through the point (0,1)
replace x with 0 and y with 1 in the equation
1= b(0*0 -1)
1 = b*-1
solve for b: multiply both sides by -1 (or divide by -1)
-1 = b*-1*-1 or
-1 = b

now you know the equation
y= -1*(x^2 -1)

you can distribute the -1 if you like:
y= -x^2 +1

Answer 12

You explained that very well. Thank you very much.

Answer 13

you can check that it works. try the point (1,0):
0= -(1*1) +1
0 = -1 + 1
0=0
it worked.