Question

The graph of a quadratic function has a y intercept at 0,5 and a minimum at 3,-4 ?

Answer 1

`f(x) = x^2 - 6x + 5`

Explanation:

`f(x)= ax^2 + bx + c`

`5 = f(0) = a( 0^2) + b(0) + c`

`c = 5`

The minimum `y` is at `x=-b/{2a}.`

`-b/{2a} = 3`

`b = -6a`

`(3,-4)` is on the curve:

`-4 = f(3) = a(3)^2 + (-6a) (3) + 5`

`-9 = -9 a`

`a = 1`

`b = -6a = -6`

`f(x) = x^2 - 6x + 5`

Check: `f(0)=5 quad sqrt`

Completing the square,

`f(x) = (x^2 - 6x + 9) -9 + 5 = (x- 3)^2 -4` so `(3,-4)` is the vertex.`quad sqrt`

Answer 2

`y=(x-3)^2-4`

Explanation:

Assuming that the equation of such quadratic graph is requested:

`y=a(x-h)^2+k` => Equation of parabola in vertex form where:

`(h, k)` is the vertex, for `a > 0` the parabola opens up which

makes the vertex the minimum, so in this case `(3, -4)` is the

vertex then:

`y=a(x-3)^2-4` => the `y` intercept is at: `(0, 5)`:

`5=a(0-3)^2-4` => solving for `a`:

`5=9a-4`

`9=9a`

`a=1`

Thus the equation of the graph is:

`y=(x-3)^2-4`