Answer:
[tex]2\,(x-1)^2=4[/tex]
which is the first option in the list of possible answers.
Step-by-step explanation:
Recall that the minimum of a parabola generated by a quadratic expression is at the vertex of the parabola, and the formula for the vertex of a quadratic of the general form:
[tex]y=ax^2+bx+c[/tex]
is at [tex]x_{vertex}=\frac{-b}{2\,a}[/tex]
For our case, where [tex]a=2\,\,, b=-4\,\,,\,\,and \,\,c=-2[/tex] we have:
[tex]x_{vertex}=\frac{-b}{2\,a}\\x_{vertex}=\frac{4}{2\,*\,2}\\x_{vertex}=1[/tex]
And when x = 1, the value of "y" is:
[tex]y(x)=2x^2-4x-2\\y(1)=2(1)^2-4(1)-2\\y(1)=2-6\\y(1)=-4\\y_{vertex}=-4[/tex]
Recall now that we can write the quadratic in what is called: "vertex form" using the coordinates [tex](x_{vertex},y_{vertex)[/tex]of the vertex as follows:
[tex]y-y_{vertex}=a\,(x-x_{vertex})^2[/tex]
Then, for our case:
[tex]y-(-4)=2\,(x-1)^2\\y=2\,(x-1)^2-4[/tex]
Then, for the quadratic equal to zero as requested in the problem, we have:
[tex]y=2\,(x-1)^2-4=0\\2\,(x-1)^2-4=0\\2\,(x-1)^2=4[/tex]
Answer:
the guy above me is correct, I go to F L V S
Step-by-step explanation:
