Answer:
To find critical points, take the first derivative and set it equal to zero:
f(x) = -2x^2 + 4x + 5
f'(x) = -4x + 4
-4x+4 = 0
-4x = -4
x = 1
Critical point at x = 1
Alternatively, if you mean zeros, or where the x intersects, you can use the quadratic equation.
Answer:
x=27
Step-by-step explanation:
1. x-2=25 so add 2 to both sides x=25+2
2. 25+2= 27 so, the answer is x=27 Can you make me Brainliest
Answer:
x=√−16n−28 or x=−√−16n−28
Step-by-step explanation:
Step 1: Add -16n to both sides.
x2+16n+28+−16n=0+−16n
x2+28=−16n
Step 2: Add -28 to both sides.
x2+28+−28=−16n+−28
x2=−16n−28
Step 3: Take square root.
x=√−16n−28 or x=−√−16n−28
The answer is 32x^4-24x^3+40x^2-36x-12
