Finding the vertex of the function P(x) = 59x - 3x^2 - 125?

Question

If I need to find the highest value of x, which i suppose is the vertex of the equation, i will have to solve this equation using completing the square method. I've tried but i am stuck at

-3(x² - 16x) + 11x - 125 , does it work this way?

Answer 1

y = -3x^2 + 59x - 125

Ax + by + c

x-value of vertex = -2b/a

-2*59/6

-118 / 6

-59/3 is the x-value of the vertex

Plug this value back in to find the y-value of the vertex

y = -3(59/3)^2 + 59(-59/3) - 125

Thank goodness for calculators

y value for vertex is -125

(-59/3 , -125)

Answer 2

No it doesnt work the way u did.

Let's complete the square:

P(x) = -3x^2 + 59x - 125

P(x) = -3 (x^2 - 59/3 x + 125/3)

P(x) = -3 [ (x - 59/6) ^2 - 59^2 / 6^2 + 125/3 ]

P(x) = -3(x - 59/6)^2 -3 ( 3481 -1500) / 36 =

= -3(x - 59/6)^2 -1981/12

The vertex is at x = 59/6

Answer 3

f(x) = y = -3x^2 + 59x -125

The vertex occurs when f '(x) = dy/dx = 0 = -6x + 59 or x = 59/6

So, f(59/6) = -3(59/6)^2 + 59(59/6) -125 = 165.0833333

Therfore, the coordinate is: (59/6, 1981/12)

Answer 4

for 59-6x=0 obtain the max of P(x)

x=59/6

P(59/6)=.....

Answer 5

x-value of vertex is actually = -b/2a = 59/6



p(x) = -3x^2 + 59x - 125

p(x) = -3 (x^2 - (59/3)x +(59/6)^2) - 125 + 3(59/6)^2

Note that I added (59/6)^2 inside the brackets (which is multiplied by -3 for a total of -3(59/6)^2) Therefore I must also subtract by the same amount outside the bracket:

-(-3(59/6)^2) = +3(59/6)^2

Now we have a square inside the bracket, and we can rewrite it as

p(x) = -3 (x^2 - (59/3)x +(59/6)^2) - 125 + 3(59/6)^2

p(x) = -3 (x - 59/6)^2 - 125 + 3481/12

p(x) = -3 (x - 59/6)^2 + 1981/12

Vertex : x-value = 59/6

p(59/6) = -3(59/6 - 59/6) + 1981/12

p(59/6) = 0 + 1981/12 = 165.083

Vertex: (59/6, 1981/12) = (9.83, 165.083)