Question

How to graph a parabola `y-2=-1/16(x-1)^2`?

Answer 1

One thing we can do is expand this function.

But first, let's just isolate `y` as follows: `y=-(1/16)(x-1)^2+2`

Now, let's expand your product of differences:

`y=-(x^2-2x+1)/16+2=(x^2-2x+33)/16`

`y=-x^2/16+x/8-2.0625`

Using Bhaskara we can find its roots:

`((-1/8)+-sqrt((1/8)^2-4(-1/16)(-33/16)))/-(2/16)`

`((-1/8)+-sqrt(1/64-33/64))/-2/16`

Ends up our `Delta` is negative, so there are no Real roots, which means this function does not cross the axis `x`.

Alternatively, we can find two other pieces of information:

Where the function crosses the axis `y`, when, by definition, `x=0`:

`y=0+0-2.0625`
`y=2.0625` is our intercept.

Now, we can find the coordinates for our vertex `(x_v,y_v)`

`x_v=-b/(2a)=(-1/8)/-(2/16)=1`
`y_v=-Delta/(4a)=1/2/-1/4=-2`

Vertex = (`1,-2`)

Finally, as the coeficient of your `x^2` is negative, you know that this is a parabola where the vertex indicates the point of maximum.

graph{-(x^2)/16+(x/8)-2.065 [-10, 10, -5, 5]}