Question

What is the vertex form of `5y = -9x^2-4x+2`?

Answer 1

`y = -9/5(x+2/9)^2+22/45`

Explanation:

A quadratic function of the form `y=ax^2+bx+c` in vertex form is given by:

`y=a(x-h)^2+k` where `(h,k)` is the vertex of the parabola.

The vertex is the point at which the parabola intersects its axis of symmetry. The axis of symmetry occurs where `x=(-b)/(2a)`

In our example: `5y=-9x^2-4x+2`

`:. y=-9/5x^2-4/5x+2/5`

Hence, `a=-9/5, b=-4/5, c=2/5`

At the axis of symmetry `x=(-(-4/5))/(2*(-9/5))`

`=-4/(2*9) = -2/9 approx -0.222`

(This is the `x-`component of the vertex, `h`)

So, `y` at the vertex is `y(-2/9)`

`= -9/5(-2/9)^2 - 4/5(-2/9) +2/5`

`= -4/(5*9) + (4*2)/(5*9) + 2/5`

`= (-4+8+18)/45 = 22/45 approx 0.489`

(This is the `y-`component of the vertex, `k`)

Hence, the quadratic in vertex form is:

`y = -9/5(x+2/9)^2+22/45`

We can see the vertex on the graph of `y` below.

graph{-9/5x^2-4/5x+2/5 [-3.592, 3.336, -2.463, 1.002]}