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

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Answer 1 · 2017-11-24T14:16:36

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

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]}