intercepts of f(x)=x^2+1
On the horizontal axis, y = 0: solve x2 + 1 = 0
⇥Let's solve in ℝ: x2 + 1 = 0
⇥ x2 = -1
⇥In ℝ, a square is never negative: no real number has -1 as its square
⇥Solution: x ∈ ∅
On the vertical axis, x = 0: calculate f(0)
⇥ 02 + 1
⇥ⓘ Adding 0 changes nothing: 0 + 1 = 1
⇥ 1
▶Axis intercepts:   (0, 1)