Grade 11 Math Question?

Question

Determine the Maximum or Minimum value for each quadratic function:

b) f(x)=2 (x-4) (x+6)
and
c) f(x)= -2x^2 +10x

Please show all steps and explain if possible, thank you.

Answer 1

ok
the first has a min point as it is a smily face curve as the coefficient infront of x^2 is positive
therefore the second is haa max.

use x part of the vertex = -b/2a when your equation is in its full quadratic form

(x)=2 (x-4) (x+6) = 2 ( x^2 +2x-24)
= 2x^2 +4x-48
so a = 2 b = 4
now -b/2a = -4 / 2 * 2 = -1 x=-1 now plug that into the original equation
y=2 (x-4) (x+6)
= 2(-1 -4) ( -1+6)
=2*-5*5
-50
min point(-1,-50)


similarly for the second one x= -b/2a= - 10 / 2* -2 = 5 / 2
and y= 25/2
max ( 5/2, 25/2)
there are many different ways of finging the max/ min point- in algebra 1 you were taught how to factor and find x interecepts- well the max/ min point is half way between them and then you plug in as shown above.

Answer 2

b) f(x)=2 (x-4) (x+6)
minimum of f(x) = f[(4-6)/2] = f(-1) = 2(-5)(5) = -50

c) f(x)= -2x^2 +10x = -2x(x-5)
maximum of f(x) = f(5/2) = -5(-5/2) = 25/2
-------
Ideas: Use symmetry. The extreme value of f(x) must be the middle points of two zeros.

Answer 3

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Answer 4

b) f(x)=2 (x-4) (x+6) = 2x^2+4x -48
f '(x) = 4x+4 = 0, so x = -1
f(-1) = 2 -4 -48 = -50 = min because x^2 term is positive

c) f(x)= -2x^2 +10x
f '(x0 = -4x+10 = 0, so x = 5/2
f(5/2) = -25/2 + 25 = 25/2 = 12.5 = max because x^2 term is negative.

Answer 5

b) f'(x)= 2(x+6) +2(x-4) = 4x +4 =0 , x=-1 , f(x) = -50
f''(x) = 4 >0 , (-1,-50 ) is minimum for f(x)
c) f'(x)= -4x +10 =0 , x = 2.5 ,y = 12.5
f"(x) = -4 < 0 , ( 2.5 ,12.5) is maximum for f(x)

Answer 6

b) f(x) = 2x^2 + 4x - 48

f'(x) = 4x + 4

4x + 4 = 0

4x = -4

x = -1 (this function only has a minimum).

f(-1) = 2 - 4 - 48

f(-1) = -50

c) f(x) = -2x^2 + 10x

f'(x) = -4x + 10

-4x + 10 = 0

4x = 10

x = 5/2 (this function has a maximum).

f(5/2) = -25/2 + 25

f(5/2) = 25/2
________________________

Or if you don't know calculus...

Your maximum/minimum is always the vertex.

V = (-b/2a , [(-b^2)/4a]+C)

-b/2a is x-coordinate of vertex

-b^2 / 4a + C is y-coordinate of vertex