f(x)(3,-3)=x^2+7

解答

f (x) (3, - 3) = x^{2} + 7

(h, k) = (0, - \frac{- 7}{3 , - 3}), p = \frac{1}{4 ( \frac{1}{( 3 , - 3 )} )}

求解步骤

y (3, - 3) = x^{2} + 7

将

y (3, - 3) = x^{2} + 7

改写为标准形式:

4 \cdot \frac{1}{4 ( \frac{1}{( 3 , - 3 )} )} (y - (- \frac{- 7}{3 , - 3})) = (x - 0)^{2}

因此抛物线性质是:

(h, k) = (0, - \frac{- 7}{3 , - 3}), p = \frac{1}{4 ( \frac{1}{( 3 , - 3 )} )}

(x - 5)^{2} + (y - 1) = 4

\circ f (x) = (5 x - 2^{2}) (6 x + 1)

(y + 5)^{2} = - 5 (x - 9)

(y - 3)^{2} = 8 x

x^{2} + 4 x - 12 y - 56 = 0