卡在這題?用 AI 解開它
Algebra · real student question

求直線 y = −4x + 2.5 與曲線 y = 3.34x/(1 + 2.34x) 的交點。

Question

求下列兩者的交點

y=−4x+2.5與y=3.34x1+2.34x.y=-4x+2.5\qquad\text{與}\qquad y=\frac{3.34x}{1+2.34x}.

Step-by-step solution

  1. 把兩個 yy 的表示式劃上等號。 在交點處兩個式子給出相同的高度,所以

    −4x+2.5=3.34x1+2.34x.-4x+2.5=\frac{3.34x}{1+2.34x}.

    記下限制 x≠−12.34≈−0.4274x\ne-\frac{1}{2.34}\approx-0.4274,分式曲線在那裡有鉛直漸近線 —— 任何落在那裡的根都必須捨棄。

  2. 去分母。 兩邊同乘 1+2.34x1+2.34x:

    (−4x+2.5)(1+2.34x)=3.34x.(-4x+2.5)(1+2.34x)=3.34x.

    把左邊逐項展開:

    −4x−9.36x2+2.5+5.85x=−9.36x2+1.85x+2.5.-4x-9.36x^{2}+2.5+5.85x=-9.36x^{2}+1.85x+2.5.

    交叉項是 −4x⋅2.34x=−9.36x2-4x\cdot2.34x=-9.36x^{2} 與 2.5⋅2.34x=5.85x2.5\cdot2.34x=5.85x;把 −4x+5.85x-4x+5.85x 合併就得到 1.85x1.85x。

  3. 整理成標準的二次式。 把 3.34x3.34x 移到左邊:

    −9.36x2+1.85x−3.34x+2.5=0  ⟹  −9.36x2−1.49x+2.5=0,-9.36x^{2}+1.85x-3.34x+2.5=0\;\Longrightarrow\;-9.36x^{2}-1.49x+2.5=0,

    再乘以 −1-1 讓首項係數為正:

    9.36x2+1.49x−2.5=0.9.36x^{2}+1.49x-2.5=0.

  4. 套公式解。 取 a=9.36a=9.36、b=1.49b=1.49、c=−2.5c=-2.5:

    Δ=1.492−4(9.36)(−2.5)=2.2201+93.6=95.8201,Δ=9.788774,\Delta=1.49^{2}-4(9.36)(-2.5)=2.2201+93.6=95.8201,\qquad\sqrt{\Delta}=9.788774,

    x=−1.49±9.78877418.72 ⟹ x≈0.4433106 或 x≈−0.6024986.x=\frac{-1.49\pm9.788774}{18.72}\ \Longrightarrow\ x\approx0.4433106\ \text{或}\ x\approx-0.6024986.

    兩者都不等於要排除的 −0.4274-0.4274,所以都保留。

  5. 求出 yy 值,並代回分式方程式檢驗。 用直線 y=−4x+2.5y=-4x+2.5:

    x=0.4433106⇒y=0.726758,x=−0.6024986⇒y=4.909994.x=0.4433106\Rightarrow y=0.726758,\qquad x=-0.6024986\Rightarrow y=4.909994.

    改代入曲線:3.34(0.4433106)1+2.34(0.4433106)=0.726758\frac{3.34(0.4433106)}{1+2.34(0.4433106)}=0.726758 ✓、3.34(−0.6024986)1+2.34(−0.6024986)=4.909994\frac{3.34(-0.6024986)}{1+2.34(-0.6024986)}=4.909994 ✓。拿另一個式子(而不是用來算 yy 的那一個)來檢查,這才算是真正的驗算。

Answer

(x,y)≈(0.4433, 0.7268)與(x,y)≈(−0.6025, 4.9100)(x,y)\approx(0.4433,\ 0.7268)\quad\text{與}\quad(x,y)\approx(-0.6025,\ 4.9100)

Need to solve a different problem like this? Open the solver →