Question

How do you find the intercept and vertex of `y= -3(x-1)²-1`?

Answer 1

Vertex`->(x,y)=(1,-1)`
`y_("intercept")=-4`

No `x_("intercept") in RR`

Explanation:

`color(blue)("Determine the vertex")`

This is the Vertex Form of a quadratic equation so you can virtually directly read off the coordinates of the vertex.

`y=-3(xcolor(red)(-1))^2color(blue)(-1)`

`x_("vertex")=(-1)xx color(red)((-1)) = +1`
`y_("vertex")=color(blue)(-1)`

Vertex`->(x,y)=(1,-1)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the y intercept")`

Set `x=0` giving

`y_("intercept")=-3(0-1)^2-1" "=" "-4`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the x intercept")`

Set `y=0` giving

`0=-3(x-1)^2-1" " larr` add 1 to both sides

`=>1=-3(x-1)^2" "larr`multiply both sides by (-1)

`=>-1=+3(x-1)^2" "larr`divide both sides by 3

`=>-1/3=(x-1)^2" "larr` square root both sides

`=>+-sqrt(-1/3)=x-1" "larr` add 1 to both sides

`x=1+-sqrt(-1/3)" " rarr x in CC`

As the determinant is negative the curve does not cross the x-axis nor is the axis a tangent to the curve.

Thus the roots are in the number range of 'Complex Numbers'
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)(" The question lists intercepts as 'singular' thus the x-axis roots are not required")`