# How do you find the vertex and intercepts for #y = x^2 – 4x + 9#?

##### 1 Answer

**vertex: (2, 5)**

**x-intercept: none**

**y-intercept: (0, 9)**

#### Explanation:

The equation is a quadratic equation in standard form, or

The **vertex** is the **minimum or maximum point of a parabola** . To find the

We know that

To find the

Simplify:

Therefore, **the vertex is at #(2, 5)#**

Now to find the intercepts.

The ** #x#-intercept is the value of #x# when #y# equals to zero**.

To find it, just plug in

Since we cannot factor

We already know that

Now simplify:

Since we cannot do a square root of a negative number (it becomes imaginary), that means there are **no #x#-intercepts** .

The ** #y#-intercept is the value of #y# when #x# equals to zero**.

To find it, just plug in

Simplify:

Now we write it as a coordinate, so it becomes

In summary,

**vertex: (2, 5)**

**x-intercept: none**

**y-intercept: (0, 9)**

Here is a graph of this quadratic equation:

As you can see, there is no

For another explanation/example of finding the vertex and intercepts of a standard equation, feel free to watch this video:

Hope this helps!