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What are the extreme values of 2x2-8x plus 9?
Given: f(x) = 2x2 - 8x + 9 This function describes a parabola, which only has one extreme value. To find that value, you can take the derivative of the function, and use that to find the point at which it has a slope of zero: f'(x) = 4x - 8 if f'(x) = 0, then: 0 = 4x - 8 ∴4x = 8 ∴ x = 2 Now we work out the corresponding value for f(2): f(2) = 2(2)2 - 8(2) + 9 = 8 - 16 + 9 = 1 So the extreme value for this function is 1, which occurs at the point (2, 1)
Factor completely -use 2 methods- 4x squared plus 4x minus 8?
4x2+4x-8 4(x2+x-2) and (4x+8)(x-1)
Which answer is a solution to 4x-8-2 0 or 2?
4x-8-2 can be more easily written as 4x-10. The solution to this equation would be 5/2
What is parenthesis 4x minus 8 parenthesis to the second power?
(4x-8)2 = (4x-8)(4x-8) = 16x2-32x-32x+64 = 16x2-64x+64
What is the y-intercept 4x- 2y 8?
4x - 2y = 8 Rearrange into the 'straight line ' form of y = mx + c Hence -2y = -4x + 8 y = 2x - 4 The 'c' value is the y-intercept, which '-4' . The steepness/gradient of the slope is '2'.
