y=ax^2 + bx + c
find a b and c such that the graph passes through, (-1,1) (0,1) (1,2)

Question

owlfred · Answer

Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

amistre64 · Answer

i gotta program for that ....

anonymous · Answer

16lb=_oz

amistre64 · Answer

attachment

amistre64 · Answer

0.5x^2  + 0.5x  + 1

amistre64 · Answer

theres a longer complicateder way; but I just programed the HTML to do it for me..

anonymous · Answer

substitue the points in the equations , form realtionn between a,b and c. solve it 
u get c=1,b=.....oh! amistre gav it for ya :)

amistre64 · Answer

(-1,1) (0,1) (1,2)

$$(-1)^2a-1b+c=1$$
$$(0)^2a+0b+c=1$$
$$(1)^2a+1b+c=2$$
is one way to go at it; the other is to see that you have 2 values for x that have the same y value

amistre64 · Answer

so your vertex is between -1 and 0;  along the x = -1/2 line

amistre64 · Answer

and your y intercept is (0,1) so if anyting c=1

anonymous · Answer

using simple substitution for each equation yields:
1= a(-1)^2 +(-1)b+c = a-b+c
1=a(0)^2+(0)b + c = c
2= a(1)^2 = b(1) +c = a+b+c
substitute eq. 2 in to eq.1
1=a-b+1 
a=b
now put eq2 and eq 1 into eq3:
2=2a+1 so that a=1/2 = b
so... y=1/2x^2 +1/2x+1