# How do you find the max or minimum of #f(x)=4x-x^2+1#?

##### 2 Answers

#### Explanation:

#f(x)=-x^2+4x+1larrcolor(blue)"in standard form"#

#"with "a=-1,b=4" and "c=1#

#• " if "a>0" then f(x) has a minimum "uuu#

#• " if "a<0" then f(x) has a maximum "nnn#

#"here "a=-1<0rArr"f(x) has a maximum"#

#"the max/min is at the vertex of the parabola"#

#"to find the vertex use "color(blue)"completing the square"#

#rArrf(x)=-(x^2-4x-1)#

#color(white)(xxxxxx)=-(x^2+(-2)x+4-4-1)#

#color(white)(xxxxxx)=-(x-2)^2+5#

#rArr"vertex at "(2,5)#

#rArr"maximum at "(2,5)#

graph{(y+x^2-4x-1)((x-2)^2+(y-5)^2-0.04)=0 [-10, 10, -5, 5]}

**Maximum is at ** :

#### Explanation:

**Given:**

**The Standard Form of a Quadratic Equation is: **

If

the **y coordinate value of the vertex** represents a **Minimum**

If

the **y coordinate value of the vertex** represents a **Maximum**

Hence, we have a **Maximum** for our problem.

Also, remember that

**Vertex: **

From .... (1) we observe that

**x coordinate of the vertex: **

Substitute the value of the x coordinate in (1) to get the y coordinate value

**Vertex:**

Hence,

**Maximum is at ** :

Analyze the graph below to understand the behavior of the quadratic:

Hope you find this solution useful.