Turning Point?

Question

I need help! My math problem says to find the turning point of the 2 functions the first one is y=4x^2 and the second one is y=x^2-x=6 Can anyone help me. Thanks

Answer 1

At the turning points, the gradient of the tangent to the curve is zero, so dy/dx = 0.

y = 4x^2
dy/dx = 8x
8x = 0
x = 0
Sub. x=0 into y=4x^2
y=0
Turning point is (0,0).

y = x^2-x+6
dy/dx = 2x-1
2x-1 = 0
2x = 1
x = 1/2 or 0.5
Sub x=1/2 into y = x^2-x+6
y = (1/2)^2 - (1/2) + 6
y = 1/4 - 1/2 + 6
y = 5 3/4 or 5.75
Turning point is (0.5, 5.75).

Answer 2

to find turning points you
1. differentitate the function
2. set the derivative equal to zero i.e. f ' (x)=0
3. solve for x
4. sub this value of x into the original function f(x) to find the value of y for the turning point.

so for y=4x^2

f ' (x) = dy/dx = 8x

f ' (x)=0 for turning points
8x=0 ===> x=0

when x=0 , sub into y=4x^2
y= 4(0)^2 = 0

.: turning point is (0,0)

Answer 3

Question 1
f(x) = 4x²
f ` (x) = 8x = 0 for turning point
x = 0 for turning point.
(0 , 0) is a turning point
f "(x) = 8 is + ve thus (0,0) is a MINIMUM turning point.

Question 2
f (x) = x² - x - 6 is equation?
f ` (x) = 2x - 1 = 0 for turning point.
x = 1 / 2 for turning point.
f "(x) = 2 is + ve (minimum turning point)
((1/2),(- 25/4)) is a minimum turning point.

Answer 4

If it's calculus then the turning point is when dy/dx=0