Is this question talking about the imaginary roots?

reshown

I assume the function crosses the y-axis twice but doesn’t cross the x-axis.

The following function is a tilted parabola (135° tilt):

2x+2y=(y-x-5)²

When x=0 (y-axis),

2y=(y-5)²,

y²-12y=-25,

y²-12y+36=11,

y=6±√11. Two y-intercepts.

When y=0 (x-axis),

2x=(x+5)²,

x²+8x+25=0, which has complex roots so does not intersect the x-axis.

But how do we know the function is a parabola? See below for transformation of a parabola.

RELOCATION (TRANSLATION) AND ROTATION

Consider the point (x₀,y₀) in the Cartesian plane x₀-y₀.

If the plane’s origin is moved to (a,b) then:

x₀→x₀+a, y₀→y₀+b.

We can write this as x₁=x₀+a, y₁+b and the new plane as x₁-y₁.

For example, (0,0)→(a,b). This means that, relative to the original plane, the origin of the new plane is at (a,b).

Now we consider rotating the new plane about its origin.

We can consider the point (x₁,y₁) in polar coordinates:

x₁=rcosθ, y₁=rsinθ

Rotating the point does not change r but does change θ. If ρ is the angle of rotation such that θ→θ+ρ, then:

x₂=rcos(θ+ρ)=rcosθcosρ-rsinθsinρ=x₁cosρ-y₁sinρ,

y₂=rsin(θ+ρ)=rsinθcosρ+rcosθsinρ=y₁cosρ+x₁sinρ.

x₂=(x₀+a)cosρ-(y₀+b)sinρ,

y₂=(y₀+b)cosρ+(x₀+a)sinρ.

From these we calculate x₀ and y₀:

x₂cosρ=(x₀+a)cos²ρ-(y₀+b)sinρcosρ,

y₂sinρ=(y₀+b)sinρcosρ+(x₀+a)sin²ρ,

x₂cosρ+y₂sinρ=(x₀+a)cos²ρ+(x₀+a)sin²ρ=x₀+a.

Therefore x₀=x₂cosρ+y₂sinρ-a.

Similarly y₀=y₂cosρ-x₂sinρ-b.

If y₀=px₀² (basic parabola) then:

y₂cosρ-x₂sinρ-b=p(x₂cosρ+y₂sinρ-a)².

We can write this in general terms:

ycosρ-xsinρ-b=p(xcosρ+ysinρ-a)².

a, b, p and ρ are arbitrary constants so let p=cosρ, ρ=135°, a=5sinρ, b=0, then:

p=cosρ=-sinρ=-1/√2, and:

-(x+y)/√2=-(1/√2)(-x+y-5)/2,

2(x+y)=(y-x-5)² which must be a parabola.

by Top Rated User (840k points)