Quadratic function

Question

Write an equation for the quadratic function which passes through the point (0,-12) and has x-intercepts at x=-6 and x=2

Answer 1

Let y=ax² +bx +c and solve for the coefficients a,b,c using the given information. When x=0 y=-12 which gives c=-12 and for (-6,0) and (2,0) 0=36a -6b -12 and 0= 4a +2b -12. Use these two equations to eliminate either a or b. To find a multiply the second equation by 3 to 0= 12a + 6b -12 and add to the first one. this gives 48a =48 and a=1 uisng this in one of the original equations gives b=4 and the function is y=x² +4x -12.

Answer 2

y=ax²+bx+c

(0,-12) gives -12=c

so now we have y=ax²+bx-12

(-6,0) gives 0=a(-6)²+b(-6)-12

36a-6b=12

6a-b=2

(2,0) gives 0=a(2)²+b(2)-12

4a+2b=12

2a+b=6

Add the two equations together:

2a+b+6a-b=6+2

8a=8

a=1

b=6-2a=6-2(1)=4

y=x²+4x-12

check: if x=0, y=-12 OK

if x=-6, y=36-24-12=0 OK

if x=2, y=4+8-12=0 OK