Answer:
1) f(x) = 0.5*(x + 3)*(x - 4)
2) f(x) = -2*x^2 - 2*x + 12
Explanation:
Wen we have a quadratic function like:
f(x) = a*x^2 + b*x + c*x
And it has roots at x1 and x2, we can write the function in the factored form as:
f(x) = a*(x - x1)*(x - x2)
then:
1) we have x-intercepts at: (-3, 0) and (4, 0) (then the roots of this function are x = -3 and x =4) and we know that f(1) = -6
we have:
x1 = -3
x2 = 4
then we can write f(x) as:
f(x) = a*(x - (-3))*(x - 4)
f(x) = a*(x + 3)*(x - 4)
Where a is a real number.
and now we can use the fact that f(1) = -6
then:
f(1) = a*(1 + 3)*(1 - 4) = -6
a*4*(-3) = -6
a*-12 = -6
a = -6/-12 = 1/2 = 0.5
Then the function is:
f(x) = 0.5*(x + 3)*(x - 4)
2) Now we have roots x = 2 and x = .3
Then:
x1 = 2
x2 = -3
Then this function is something like:
f(x) = a*(x - 2)*(x - (-3))
f(x) = a*(x - 2)*(x + 3)
Now we know that f(0) = 12.
then:
f(0) = a*(0 - 2)*(0 + 3) = 12
a*(-2)*3 = 12
a*(-6) = 12
a = 12/-6 = -2
f(x) = -2*(x - 2)*(x + 3)
And we do not want this one written in factored form, so we can just distribute the multiplications to et:
f(x) = -2*(x - 2)*(x + 3) = (-2*x + 4)*(x + 3) = -2*x^2 + 4*x - 6*x + 12
f(x) = -2*x^2 - 2*x + 12