Question

Let f(x)=ax^2+bx+c for some value of a, b and c. f intersects the x-axis when x=2 or x=4, and f(1)=6. Find the values of a, b, and c.

Answer 1

so you have 3 points given: (2,0), (4,0) and (1,6)
sub in (2,0) ---> 4a + 2b + c = 0 **
sub in (4,0) --> 16a + 4b + c = 0 ***
subtract them: 12a + 2b = 0 or b = -6a
sub in (1,6) --> a + b + c = 6 ***
subtract *** from **
4a + 2b + c = 0
a + b + c = 6
3a + b = -6, but b = -6a
3a - 6a = -6
a = 2 , then b = -12
back in ****
a+b+c=6
2 - 12 + c = 6
c = 16

f(x) = 2x^2 - 12x + 16

verification: https://www.wolframalpha.com/input/?i=f%28x%29+%3D+2x%5E2+-+12x+%2B+16

Answer 2

for x=2 ... 0 = 4a + 2b + c

for x=4 ... 0 = 16a + 4b + c

for x=1 ... 6 = a + b + c

solve the system for a , b , and c

Answer 3

The roots are x=2 and x=4, so
y = a(x-2)(x-4)
Since y(1) = 6,
a(1-2)(1-4) = 6
a = 2
So, y = 2(x-2)(x-4) = 2x^2 - 12x + 16

Answer 4

oobleck, your solution is neater than mine.
Sometimes, one doesn't see the forest for the trees.