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

2 Answers
Jan 14, 2018

#"maximum at "(2,5)#

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]}

Jan 14, 2018

Maximum is at : #color(blue)[(2,5)#

Explanation:

Given:

#color(red)(y = f(x) = 4x - x^2 +1)#

#rArr f(x) = - x^2 + 4x +1#

#Let " " f(x) = - x^2 + 4x +1 = 0# ...(1)

The Standard Form of a Quadratic Equation is:

#f(x) = ax^2 + bx + c = 0#

If #a > 0# then

the y coordinate value of the vertex represents a Minimum

If #a < 0# then

the y coordinate value of the vertex represents a Maximum

Hence, we have a Maximum for our problem.

Also, remember that

Vertex:

#[ ((-b)/(2a)), f((-b)/(2a))]#

From .... (1) we observe that

#color(blue)(a = -1; b = 4 and c = 1)#

x coordinate of the vertex:

#(-b)/(2a) = (-4)/(2(-1)) = ((-4)/-2) = 2#

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

#y = - x^2 + 4x +1#

#rArr -(2)^2 + 4(2) + 1#

#rArr -4 + 8 + 1#

#rArr 5#

#:.# Vertex: #color(blue)[(2,5)#

Hence,

Maximum is at : #color(blue)[(2,5)#

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

enter image source here

Hope you find this solution useful.