So since the vertex falls onto the axis of symmetry, we can just solve for that to get the x-coordinate of both equations. The equation for the axis of symmetry is $x=\frac{-b}{2a}$ , with b = x coefficient and a = x^2 coefficient. Our equations can be solved as such:

y = 2x^2 − 4x + 12: $x=\frac{4}{2*2}=\frac{4}{4}=1$

y = 4x^2 + 8x + 3: $x=\frac{-8}{2*4}=\frac{-8}{8}=-1$

In short, the vertex x-coordinate's of y = 2x^2 − 4x + 12 is 1 while the vertex's x-coordinate of y = 4x^2 + 8x + 3 is -1.

6 0

3 years ago

How would I solve <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20x-%5Cfrac%7B5%7D%7B3%7D%20%3D-%5Cfrac%7B1%7D%7B2%7D%

sdas [7]

Multiply both sides of the equation by 12. Move the variables to the left-hand side and change its sign. Move the constant to the left-hand side and change its sign. Collect like terms. Add the numbers. Divide both sides of the equation by 12.

1/2 x - 5= - 1/2 x + 19/4
6x -20= - 6x + 57
6x + 6x= 57 + 20
12x= 57 + 20
12x= 77
ANSWER
x = 77/12
Alternative Form .
x = 6 5/12, x= 6.416

7 0

3 years ago

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