# How do you find the max or minimum of #f(x)=x-2x^2-1#?

##### 2 Answers

This parabola has a maximum point at vertex:

#### Explanation:

When you have a parabola in standard form

Here are some rules

- If
#a>0# , then vertex is a minimum - If
#a<0# , then vertex is a maximum - Vertex is given by the following formula

Vertex:

Step 1. Rewrite the equation in standard form

Step 2. Determine maximum or minimum

Step 3. Plug into the vertex formula

Vertex:

graph{-2x^2+x-1 [-2, 2, -5, 0.1]}

refer below

#### Explanation:

The given equation is Quadratic, which means it only has 1 maximum or 1 minimum point.

**1.** Re-arranging the equation,

Since the coefficient of the term **negative**, the graph of the curve has a maximum point. Conversely, if it is a *positive*, it has a minimum point.

graph{-2x^2+x-1 [-4.896, 5.104, -4.66, 0.34]}

The curve is shown in this graph.

**2.** At the turning point, the **gradient** of the function is **zero**.

This means

Differentiating the function,

#-4x=-1#

#color(blue)(x=1/4)#

**3.** Most questions require you to prove/determine the turning point of the function is a maximum/minimum point.

To prove that its **turning point** is a maximum, you have to use the **Second Derivative Test**.

A maximum point will have

(a minimum point will have f''(x)>0)

Differentiating f'(x),

Since **maximum**

**4.** To find the coordinates of the **maximum**, substitute the

when

#y=-7/8#

**5.** Coordinates of the maximum is