help please I just want to know how to do it

Question

Sketch each of the following parabolas, clearly showing the axis intercepts and the turning point:

1. y=2x²−4x 
2. y=−2x²−3x
3. y=x²−5x+4
4. y=2x²−x−1
5. y=−x²−5x−6

(Don't need to sketch it, just need to how to find the tp)

Answer 1

y=2x^2 -4x

y'(x)= dy/dx = 4x -4 = 0

x-1 = 0

x=1

y = 2-4 =-2

(1,-2) is the vertex, the minimum point, the turning point where y changes direction and goes up as x increases, for x<1, y is going down hill, for x>1, y goes up. at x=1, y is for an instant not changing value

or you could rewrite y=2x^2-4x in vertex form

y=2x^2 -4x

y=2(x^2-2x)

complete the square

y = 2(x^2 -2x+1) -2

y = 2(x-1)^2 -2 vertex = (1,-2) = minimum point = turning point

y=a(x-h)^2 +k is vertex form for a quadratic equation with (h,k) = vertex = (1,-2)

h=1, k=-2

y=-2x^2 -3x

take the derivative and set equal to zero

y' = -4x-3 = 0

x =-3/4

y = -2(-3/4)^2 -3(-3/4) = -2(9/16 +9/4 = -9/8+18/8 =9/8

turning point (h,k) = (-3/4, 9/8)

or rewrite the quadratic equation in vertex form

y =-2(x^2 +3x/2)

complete the square,

y =-2(x^2 +3x/2 +(3/4)^2) + 2(3/4)^2

y=-2(x+3/4)^2 + 9/8

turning point = (-3/4,9/8)

1), 3) and 4) are upward opening parabolas

2) and 5) are downward opening parabolas