Y = X2 - 4X - 5
set to zero
X2 - 4X - 5 = 0
X2 - 4X = 5
halve the linear term ( - 4 ) then square it and add that result to both sides
X2 - 4X + 4 = 5 + 4
factor on the left and gather terms together on the right
(X - 2)2 = 9
(X - 2)2 - 9 = 0
==============vertex form
(2, - 9)
======vertex
The vertex of a parabola is the minimum or maximum value of the parabola. To find the maximum/minimum of a parabola complete the square: x² + 4x + 5 = x² + 4x + 4 - 4 + 5 = (x² + 4x + 4) + (-4 + 5) = (x + 2)² + 1 As (x + 2)² is greater than or equal to 0, the minimum value (vertex) occurs when this is zero, ie (x + 2)² = 0 → x + 2 = 0 → x = -2 As (x + 2)² = 0, the minimum value is 0 + 1 = 1. Thus the vertex of the parabola is at (-2, 1).
In ordinary mathematics, assuming that x = X and that X2 denotes x2 or x-squared, there cannot be a counterexample since the statement is TRUE. However, there are two assumptions made that could be false and so could give rise to counterexamples. 1. x is not the same as X. If, for example X = 4x then X = -20 so that X2 = 400. 2a. X2 is not X2 but X times 2. In that case X2 = -10. 2b. X2 is x2 modulo 7, for example. Then X2 = 4.
-5
Differentiate the function with respect to x: d/dx (x3 - 2x2 - 5x + 6) = 3x2 - 4x - 5 Set this derivative = 0 and solve. 3x2 - 4x - 5 = 0 implies that x = -0.7863 or 2.1196 (to 4 dp)
3x+5= 4x-10+10 +103x+15= 4x-3x -3x15=X
Example function.Y = X2 - 4X + 5set to 0X2 - 4X + 5 = 0X2 - 4X = - 5Now, halve the linear coefficient, square it and add it to both sidesX2 - 4X + 4 = - 5 + 4gather terms on the right and factor on the left(X - 2)2 = -1==============Vertex form.(2, - 1)=======Vertex.
The vertex is at (5, -5).
2x2= 4
Their sum is 4.
You can work this out by taking the derivative of the equation with respect to y, and finding where that derivative (ie. the slope of the curve's tangent) equals 0: y = -x2 + 4x - 9 ∴ dy/dx = -2x + 4 Let dy/dx = 0: 0 = -2x + 4 ∴ x = 2 Now find the corresponding y co-ordinate by plugging that value for x into the original equation: y = -(2)2 + 4(2) - 9 ∴ y = - 4 + 8 - 9 ∴ y = -5 So the vertex of that parabola is located at the point (2, -5).
Since the polynomial has three zeroes it has a degree of 3. So we have:(x - -5)[x - (2 + i)][x - (2 - i)]= (x + 5)[(x - 2) - i][(x - 2) + i]= (x + 5)[(x - 2)2 - i2]= (x + 5)[x2 - 2(x)(2) + 22 - (-1)]= (x + 5)(x2 - 4x + 4 + 1)= (x + 5)(x2 - 4x + 5)= x(x2 - 4x + 5) + 5(x2 - 4x + 5)= x3 - 4x2 + 5x + 5x2 - 20x + 25= x3 + x2 - 15x + 25Thus,P(x) = x3 + x2 - 15x + 25
The coordinates will be at the point of the turn the parabola which is its vertex.
Add 4 to both sides. To complete the square, take half the x coefficient into the bracket: (x + 4/2)2 = (x + 2)2 = x2 + 4x + 4 Comparing with the original, need to add 4 to both sides of x2 + 4x = 5: (x2 + 4x) + 4 = (5) + 4 ⇒ x2 + 4x + 4 = 9 ⇒ (x + 2)2 = 9
X2 - 4X - 5 as an expression, or equation is factorable when the question, what two factors of - 5 add to - 4 is answered ( X + 1)(X - 5) X = - 1 ------------and X = 5 ---------------
Assuming the missing operator before the final 5 is subtract, then: x2 - 4x - 5 = (x + 1) (x - 5) ⇒ x = -1 or 5. If it is add, there are no real root solution. The solution to the latter case is: x2 - 4x + 5 = (x - 2 - i)(x - 2 + i) ⇒ x = 2 + i or x = 2 - i.
f(x) = x2-4x-5 = x2-5x+x-5 = x(x-5)+1(x-5) = (x+1)(x-5) Factors of f(x) are (x+1) and (x-5).
If that's -5 the answer is (x - 5)(x + 1)