Question

How do you write a quadratic equation with Vertex: (6, 5); point: (0, -4)?

Answer 1

`y = f(x) = -(1/4)x^2 + 3x - 4`

Explanation:

We are given the following details:

Vertex is at `( 6, 5)`

One Point on the Parabola = `( 0,-4 )`

We must write a Quadratic Equation using the given details.

Standard Form = `y = f(x) = ax^2 + bx + c = 0`

Vertex Form = `y = f(x) = a( x - h ) ^ 2 + k` where Vertex `= ( h, k )`

Vertex Form is appropriate for our question. Hence we will use this form to solve the problem.

Our Vertex is at `( 6, 5)`. Using this information we can write

`y = f(x) = a( x - 6 )^2 + 5 color(red)( Equation.A)`

We are also given a Point (0, -4)

Hence, x = 0; y = -4

Substitute these values in `color(red)( Equation.A)`

`-4 = a(0 - 6)^2 + 5`

If we simplify for a, we get `a = (-)(1/4)`

So, our f(x) now can be written as

`f(x) = (-)(1/4)( x - 6)^2 + 5`

We can also rewrite this in it's simplified form as follows:

`y = f(x) = -(1/4)x^2 + 3x - 4`

graph{-(1/4)x^2 + 3x - 4 [-10, 10, -5, 5]}