Question

Using the completing-the-square method, find the vertex of the function f(x) = –2x2 + 12x + 5 and indicate whether it is a minimum or a maximum and at what point.

Maximum at (–3, 5)
Minimum at (–3, 5)
Maximum at (3, 23)
Minimum at (3, 23)

Answer 1

Answer:

Step-by-step explanation:

f(x) = –2x² + 12x + 5

     = -2 (x² - 6x - 5/2)

     = - 2 ( x² -2(3)x +9 - 9 - 5/2)

     = - 2 ( (x - 3)²-23/2)

f(x) = - 2 ( x - 3)²+23 .....vertex form

the vertex is : (3 : 23)   Maximum at (3, 23) because a = -2 and a< 0

Answer 2

Answer:

Maximum at (3, 23)

Step-by-step explanation:

$f(x) = -2x^2 + 12x + 5$

f(x)= a(x-h)^2 +k , where (h,k) is the vertex

Apply completing the square method to find vertex

$f(x) = -2x^2 + 12x + 5$

$f(x) = -2(x^2 -6x)+ 5$

Let take half of coefficient of x is -6 divide by 2 is -3

square it (-3)^2 is 9

Add and subtract 9

$f(x) = -2(x^2 -6x+9-9)+ 5$

Take out -9 and multiply by -2

$f(x) = -2(x^2 -6x+9)+18+ 5$

$f(x) = -2(x^2 -6x+9)+23$

Now factor the parenthesis part

$f(x) = -2(x-3)^2+23$

The value of h=3  and k=23

So vertex is (3,23)

The value of 'a' is -2, it means the parabola is upside down. so vertex is maximum

vertex is maximum at (3,23)