The question is as followed:
y=-3x^2 + 4x
This is how I solved it:
y=-3x^2 + 4x
y= -3 (x^2 + 4/3x)
y= -3(x^2+4/3x +4/9) + 4/3
y= -3 (x+2/3)^2 + 4/3
The answer sheet said that x = 2/3 but wouldn't it equal -2/3 because there is a -3 before the bracket?
y=-3x^2 + 4x
y= -3 (x^2 + 4/3x)You erred here, it should be
y=-3x^2 + 4x
y= -3 (x^2 - 4/3x )
oh ok
thanks so much :)
To solve the equation y = -3x^2 + 4x, we need to find the value(s) of x that make y equal to zero. This is because when y = 0, the equation is satisfied.
To determine the values of x, we can set y equal to zero and factor the equation:
0 = -3x^2 + 4x
Now, we can factor out common terms from the equation:
0 = x(-3x + 4)
To find the values of x, we can set each factor equal to zero:
x = 0 or -3x + 4 = 0
For the first factor, x = 0, which means one solution is x = 0.
For the second factor, -3x + 4 = 0, we can solve for x as follows:
-3x = -4
x = -4/-3
x = 4/3 or 1 1/3
So, the equation has two solutions: x = 0 and x = 4/3 (or x = 1 1/3).
However, it seems like there might be some confusion regarding the solution x = 2/3. In the steps you provided, it appears that you have rewritten the equation in vertex form using completing the square:
y = -3(x + 2/3)^2 + 4/3
The vertex form of a quadratic equation is given by y = a(x - h)^2 + k, where (h, k) represents the coordinates of the vertex. In this case, the vertex would be (-2/3, 4/3).
Therefore, it seems like there might be a mistake in the answer sheet, as x = 2/3 is not a solution to the original equation y = -3x^2 + 4x. The correct solutions are x = 0 and x = 4/3 (or x = 1 1/3).