# How do you find the vertex and intercepts for #y = 5(x+2)^2 + 7#?

##### 1 Answer

**Vertex of the parabola** formed by

**x-intercept: Does not exist.**

**y-intercept: **

#### Explanation:

**Standard form of a quadratic function is **

**The parabola will open up **, if the the coefficient of **positive**.

**The parabola will open down **, if the the coefficient of **negative**.

Let us consider the **quadratic function given to us:**

Using the algebraic identity

We have seen that

**Standard form of a quadratic function is **

Note that,

**To find the x-coordinate of the Vertex**, use the formula

To find the **y-coordinate of the vertex**, substitute

Hence, **Vertex ** is at

Since, the **coefficient of the #x^2# term is positive**, the

**parabola opens up**.

**x-intercept** is a point on the graph where

Solve

Subtract

Divide both sides by

Observe that

**y-intercept is the point on the graph where #x=0#**.

**-intercept is at **

Hence,

**Vertex of the parabola** formed by:

is

**x-intercept: Does not exist.**

**y-intercept: **

An image of the graph is available below: