Δf(x)=5.15*y^2-0.15

解

Δ f (x) = 5.15 \cdot y^{2} - 0.15

(h, k) = (- \frac{0.15}{Δ f}, 0), p = 0.04854 \dots Δ f

解答ステップ

Δ f x = 5.15 y^{2} - 0.15

Δ f x = 5.15 y^{2} - 0.15

を標準的な形式で書き換える:

4 \cdot 0.04854 \dots Δ f (x - (- \frac{0.15}{Δ f})) = (y - 0)^{2}

ゆえに放物線の特性は:

(h, k) = (- \frac{0.15}{Δ f}, 0), p = 0.04854 \dots Δ f