Question

How do you write a quadratic function whose graph has the given characteristics: passes through (5,2), (0,2), (8,-6)?

Answer 1

`f(x)=-1/3x+5/3x+2`

Explanation:

Let the function be `f(x)`. From the `(0,2)` and `(5,2)`, we can determine that `f(x)-2=a(x-5)(x)` where a is the leading coefficient of the first term, because if you were to shift the function down 2, it would have zeros at 0 and 5. a is negative, given that the position of the third point `(8,-6)`.
The a value is less than one.

`f(x)-2=a(x-5)(x)`
`f(x)=a(x^2-5x)+2`
`f(x)=ax^2-5ax+2`

We know `(8,-6)` is a point on `f(x)`, so plug in the point into the equation, and get a.
`-6=a(8)^2-5a(8)+2`
`-8=64a-40a`
`-8=24a`
`a=-1/3`

Now, write the equation using the fact that `a=-1/3`.
`f(x)=-1/3x^2-5(-1/3)x+2`
`f(x)=-1/3x+5/3x+2`