Question

PLLLLLSSSS Heeeellppp ill mark brainliest I promisseeeeeeee plsssss

Answer 1

Answer:

$2\,(x-1)^2=4$

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:

$y=ax^2+bx+c$

is at   $x_{vertex}=\frac{-b}{2\,a}$

For our case, where $a=2\,\,, b=-4\,\,,\,\,and \,\,c=-2$  we have:

$x_{vertex}=\frac{-b}{2\,a}\\x_{vertex}=\frac{4}{2\,*\,2}\\x_{vertex}=1$

And when x = 1, the value of "y" is:

$y(x)=2x^2-4x-2\\y(1)=2(1)^2-4(1)-2\\y(1)=2-6\\y(1)=-4\\y_{vertex}=-4$

Recall now that we can write the quadratic in what is called: "vertex form" using the coordinates $(x_{vertex},y_{vertex)$of the vertex as follows:

$y-y_{vertex}=a\,(x-x_{vertex})^2$

Then, for our case:

$y-(-4)=2\,(x-1)^2\\y=2\,(x-1)^2-4$

Then, for the quadratic equal to zero as requested in the problem, we have:

$y=2\,(x-1)^2-4=0\\2\,(x-1)^2-4=0\\2\,(x-1)^2=4$