Question

How do you find the vertex and axis of symmetry, and then graph the parabola given by: `f(x)=(x-5)^2 - 9`?

Answer 1

`(5;-9)`

Explanation:

There are 2 methods:
Method 1
Multiply out and write in standard quadratic form `y=ax^2+bx+c`
Then vertex (max/min) value occurs at point when derivative is zero, ie. when `2ax+b=0=>x=(-b)/(2a)`
So in this case, we get :
`y=f(x)=x^2-10x+16`
`therefore dy/dx=0iff2x-10=0iffx=10/2=5`
`therefore f(5)=-9=>` vertex occurs at `(5;-9)`

Method 2
The completed square form of the quadratic function is `y=a(x-p)^2+q`
In this case, `(p;q)` represents the vertex, ie p is the x value at the axis of symmetry and q is the corresponding y value.
So in this example, p = 5 and q = - 9 and so we immediately get the vertex or turning point at `(5;-9)`

Graphically :

graph{((x-5)^2)-9 [-32.47, 32.47, -16.24, 16.25]}

Answer 2

Vertex: (5,-9)
Axis of Symmetry: x=5

Explanation:

That equation is written in the vertex form `f(x)=a(x-h)^2+k`.

Vertex
The vertex is the point `(h,k)`. Take a look at the equation. You'll see that your `h` is `5`, and your `k` is `-9`. Therefore, your vertex is `(5,-9)`. Plot that on your graph.

Axis of Symmetry
In a quadratic function, the axis of symmetry is a vertical line that passes through the vertex. The axis of symmetry is written as `x=h`. Looking at your equation, `h` is equal to `5`. Therefore, the axis of symmetry is `x=5`. You don't actually have to draw this on your graph though.

Finally, when drawing the graph, create a table of `x` and `y` values. To fill it up, substitute any value to `x` and write down the corresponding value of `y`. Try to substitute easy numbers like 0, 1, and 2. After that, plot the points and draw the parabola. It should look like this:
graph{(x-5)^2-9 [-20, 20, -10, 10]}