Determine whether the given quadratic function has a minimum value or maximum value. Then find the coordinates?

it has a minimum value of course

Question Number 1 : For this function y(x)= -x^2 + 2*x - 4 , answer the following questions : A. Find the minimum/maximum point of the function ! Answer Number 1 : The equation -x^2 + 2*x - 4 = 0 is already in a*x^2+b*x+c=0 form. By matching the constant position, we can derive that the value of a = -1, b = 2, c = -4. 1A. Find the minimum/maximum point of the function ! Since the value of a = -1 is negative, the function y(x) = -x^2 + 2*x - 4 have a maximum point. Since maximum point is where the curve turn, so the formula y'(x) = 0 , can be used to find the value of x We have to find the function y'(x) first So we get y'(x) = - 2*x + 2 = 0 Which means that -2*x = -2 Which means that x = -2/-2 So we get x = 1 So the maximum point is ( x , y ) = ( 1 , y(1) ) Which is ( x , y ) = ( 1 , -3 ) So the answer is B Max (1,-3)

f(x) will have a critical point where f'(x)=0. -4x - 4 = 0, x=-1

The coordinates of the critical point are (-1, 2).

f''(-1) = -4 is <0 so the critical point is a maximum.

is this -2x^2 - 4x?

if so

f'(x) = -4x - 4

set f'(x) to 0 to find critical number(s)

-4x - 4 = 0

x = -1

use second derivative test to see if is local min or max.

f''(x) = -4

f''(-1) = -4

which means when x = -1 the function is concave down so it must be a maximum.

f(-1) = -2(-1)^2 -4(-1)

f(-1) = 2

coords are (-1,2)

f(x) = -2x² - 4x

f(x) = -2(x² + 2x)

f(x) = -2(x² + 2x + 1) + 2

f(x) = -2(x + 1)² + 2

max at (-1,2)