Question

How do you find the vertex and intercepts for `x = –2(y– 3)^2 - 2`?

Answer 1

Vertex at `(-2,3)`

`x_("intercept")=-20`

Explanation:

Given `x=-2(ycolor(blue)(-3))^2color(magenta)(-2)`

`color(blue)("Solving for vertex")`

As in completing the square for x we have the same scenario but this time in y

`"y-vertex" = (-1)xxcolor(blue)(-3) = +3`

`"x-vertex" = color(magenta)(-2)`

`=>"Vertex"->(x,y)->(-2,3)`
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`color(blue)("Solving for x intercept")`

x intercept @ y=0 giving

`x=-2(y-3)^2-2 -> x=-2(9)-2 = -20`

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`color(blue)("Solving for y intercept")`

The determinant for `y=(-b+-sqrt(b^2-4ac))/(2a)` Must be positive

`x=-2(y-3)^2-2" "->" "x=-2y^2+12y-20`

Thus the determinant is: `sqrt(12^2-4(-2)(-20)) = sqrt(-16 )`

Thus there are only Complex number roots for x=0

`color(magenta)(y_("intercept") " Does not exist")`
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