2x^2-13x+k(x)=0

解

2 x^{2} - 13 x + k (x) = 0

(h, k) = (\frac{13}{4}, \frac{169}{8}), p = - \frac{1}{8}

解答ステップ

2 x^{2} - 13 x + y = 0

2 x^{2} - 13 x + y = 0

を標準的な形式で書き換える

4 (- \frac{1}{8}) (y - \frac{169}{8}) = (x - \frac{13}{4})^{2}

ゆえに放物線の特性は:

(h, k) = (\frac{13}{4}, \frac{169}{8}), p = - \frac{1}{8}