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

Sep 25, 2015

(5;-9)

#### Explanation:

There are 2 methods:
Method 1
Multiply out and write in standard quadratic form $y = a {x}^{2} + b x + c$
Then vertex (max/min) value occurs at point when derivative is zero, ie. when $2 a x + b = 0 \implies x = \frac{- b}{2 a}$
So in this case, we get :
$y = f \left(x\right) = {x}^{2} - 10 x + 16$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 0 \iff 2 x - 10 = 0 \iff x = \frac{10}{2} = 5$
$\therefore f \left(5\right) = - 9 \implies$ vertex occurs at (5;-9)

Method 2
The completed square form of the quadratic function is $y = a {\left(x - p\right)}^{2} + q$
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]}

Sep 25, 2015

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

#### Explanation:

That equation is written in the vertex form $f \left(x\right) = a {\left(x - h\right)}^{2} + k$.

Vertex
The vertex is the point $\left(h , k\right)$. Take a look at the equation. You'll see that your $h$ is $5$, and your $k$ is $- 9$. Therefore, your vertex is $\left(5 , - 9\right)$. 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]}