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

Nov 24, 2017

$y = f \left(x\right) = - \left(\frac{1}{4}\right) {x}^{2} + 3 x - 4$

Explanation:

We are given the following details:

Vertex is at $\left(6 , 5\right)$

One Point on the Parabola = $\left(0 , - 4\right)$

We must write a Quadratic Equation using the given details.

Standard Form = $y = f \left(x\right) = a {x}^{2} + b x + c = 0$

Vertex Form = $y = f \left(x\right) = a {\left(x - h\right)}^{2} + k$ where Vertex $= \left(h , k\right)$

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

Our Vertex is at $\left(6 , 5\right)$. Using this information we can write

$y = f \left(x\right) = a {\left(x - 6\right)}^{2} + 5 \textcolor{red}{E q u a t i o n . A}$

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

Hence, x = 0; y = -4

Substitute these values in $\textcolor{red}{E q u a t i o n . A}$

$- 4 = a {\left(0 - 6\right)}^{2} + 5$

If we simplify for a, we get $a = \left(-\right) \left(\frac{1}{4}\right)$

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

$f \left(x\right) = \left(-\right) \left(\frac{1}{4}\right) {\left(x - 6\right)}^{2} + 5$

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

$y = f \left(x\right) = - \left(\frac{1}{4}\right) {x}^{2} + 3 x - 4$

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