Max and Min of Quadratic Functions?

Question

Determine, WITHOUT GRAPHING, if the given quadratic functions have a maximum or minimum value and then find the value.

(1) f(x) = -2x^2 + 12x

(2) f(x) = 4x^2 - 8x + 3

Answer 1

all quadratic eqns (ax^2 + bx + c) has either a maximum or a minimum value.
this pt is given by x = -b/2a
it is either a maximum or a minimum depending only on the value of a. if a > 0, it is a minimum and if a < 0, it is a maximum.

#1 x = -12/(2*-2) = 3. y = 18.
a = -2. it is a maximum.

#2 x = 8/(2*4) = 1. y = -1.
a = 4. it is a minimum.

Answer 2

For each, calculate the first derivative, set the result equal to zero, and solve for x. The values of x correspond to maxima and minima in the function.

f(x) = -2x² + 12x
f'(x) = -4x + 12 = 0
x = -3

then f(3) = -2(9) + 12(3) = -18+36 = 18

f''(x) = -4 < 0 so it's a maximum.

f(x) = 4x² - 8x + 3
f'(x) = 8x - 8 = 0
x = 1

then f(1) = 4-8+3=-1

f''(x) = 8 >0 so it's a minimum.

Answer 3

find the derivative:
1.) f'(x)=-4x + 12
set equal to zero
x=3
then plug this value into your derivative which gives you
-4(3) + 12=0
set up a chart
(-infinity, 3)(3, infinity)
find a value inbetween those and plug it into your derivative (I picked 0 for the left side)
-4(0) +12 =12
since 12 is positive you'll have a postive slope on the left side
pick a number on the right (I picked 4) and plug it into the derivative
-4(4) + 12=-4
since -4 is negative, you'll have a negative slope on the right side
your slopes will look like this:
/\
since they meet at the top, 3 is a maxium of the function (it has no minimum which makes sense because it's a parabola)
to find the cooridnate of the max, plug 3 into your ORIGINAL equation to get your y value

2.) repeat

Answer 4

if it has a reflection on x axis, it will have a maximum, if it has no reflection, it will have a minimum.

there are different ways to approach the value.

you can find the x-coordinate of the vertex by calculating (-b/2a) and plugging that into the function to find y-coordinate.

you can also find the derivative and set it equal to zero and solve for x. a quadratic will have only 1 critical point, so you will get only 1 answer. again, you plug the x into the function and calculate y.

the worst (in my humble opinion) method would be completing the square and rearranging the function into transformational form and reading off the vertex coordinates as HT and VT's. it works, but is long and tedious.

Answer 5

(1) f(x) = -2x^2 + 12x has maximum since it opens down.
x = -b/(2a) = 12/4 = 3
fmax = f(3) = -2(3^2) + 12(3) = 18

(2) f(x) = 4x^2 - 8x + 3 has minimum since it opens up.
x = -b/(2a) = 8/8 = 1
fmin = f(1) = 4 - 8 + 3 = -1

Answer 6

If it has a minimum it has a happy grin
If it has a maximum it has a sad mouth
if the coefficient of x^2 is positive it is happy
if the coefficient of x^2 is negative it is sad

Answer 7

y=48x^2-24x+c would have 2 intercepts while Determinant >0 i.e 576-4*40 8*c>0 576-192c>0 192c<576 c<3 the optimal cost of a quadratic cost while a>0 is infinity and minimum cost is -D/4a the place D is discriminant