Применяем формулу нахождения координаты x вершины параболы:
1 . x = -b/2a ,
2 .x = -12/2*2 = -3
Тогда 3. y = 2(-3*-3) + 12 *(-3) + 5 = 18 - 36 + 5 = -13
Ответ: Координаты вершины параболы (-3;-13)
Решение:
a^2+8a+16=(a+4)^2 Вот и всё!
4x^2+3x-16x-12+3=0
4x^2-13x-9=0
D=<span><span><span><span><span>(<span>−13</span>)^</span>2</span>−<span><span>4·4</span>·<span>(<span>−9</span>)</span></span></span>=<span>169+144</span></span>=313=17,7
</span><span>x1=-(-13)+17,7/2*4= 30,7/8=3,8
X2=-(-13)-17,7/2*4=-4,7/8=-0,6
</span>
ОДЗ : x + 1 ≥ 0 ⇒ x ≥ - 1
![14*4^{\sqrt{x+1}}+3*14^{\sqrt{x+1}}-2*49^{\sqrt{x+1}}=0\\\\\frac{14*4^{\sqrt{x+1}}}{4^{\sqrt{x+1}}}+\frac{3*14^{\sqrt{x+1}}}{4^{\sqrt{x+1}}}-\frac{2*49^{\sqrt{x+1}}}{4^{\sqrt{x+1}}}=0\\\\14+3*(\frac{7}{2})^{\sqrt{x+1}}-2*[( \frac{7}{2})^{\sqrt{x+1}}]^{2}=0\\\\(\frac{7}{2})^{\sqrt{x+1}}=m,m>0\\\\14+3m-2m^{2}=0\\\\2m^{2}-3m-14=0\\\\D=(-3)^{2}-4*2*(-14)=9+112=121=11^{2}\\\\m_{1}=\frac{3+11}{4}=\frac{7}{2}\\\\m_{2}=\frac{3-11}{4}=-2<0-neyd\\\\(\frac{7}{2})^{\sqrt{x+1}}=\frac{7}{2}](https://tex.z-dn.net/?f=14%2A4%5E%7B%5Csqrt%7Bx%2B1%7D%7D%2B3%2A14%5E%7B%5Csqrt%7Bx%2B1%7D%7D-2%2A49%5E%7B%5Csqrt%7Bx%2B1%7D%7D%3D0%5C%5C%5C%5C%5Cfrac%7B14%2A4%5E%7B%5Csqrt%7Bx%2B1%7D%7D%7D%7B4%5E%7B%5Csqrt%7Bx%2B1%7D%7D%7D%2B%5Cfrac%7B3%2A14%5E%7B%5Csqrt%7Bx%2B1%7D%7D%7D%7B4%5E%7B%5Csqrt%7Bx%2B1%7D%7D%7D-%5Cfrac%7B2%2A49%5E%7B%5Csqrt%7Bx%2B1%7D%7D%7D%7B4%5E%7B%5Csqrt%7Bx%2B1%7D%7D%7D%3D0%5C%5C%5C%5C14%2B3%2A%28%5Cfrac%7B7%7D%7B2%7D%29%5E%7B%5Csqrt%7Bx%2B1%7D%7D-2%2A%5B%28%20%5Cfrac%7B7%7D%7B2%7D%29%5E%7B%5Csqrt%7Bx%2B1%7D%7D%5D%5E%7B2%7D%3D0%5C%5C%5C%5C%28%5Cfrac%7B7%7D%7B2%7D%29%5E%7B%5Csqrt%7Bx%2B1%7D%7D%3Dm%2Cm%3E0%5C%5C%5C%5C14%2B3m-2m%5E%7B2%7D%3D0%5C%5C%5C%5C2m%5E%7B2%7D-3m-14%3D0%5C%5C%5C%5CD%3D%28-3%29%5E%7B2%7D-4%2A2%2A%28-14%29%3D9%2B112%3D121%3D11%5E%7B2%7D%5C%5C%5C%5Cm_%7B1%7D%3D%5Cfrac%7B3%2B11%7D%7B4%7D%3D%5Cfrac%7B7%7D%7B2%7D%5C%5C%5C%5Cm_%7B2%7D%3D%5Cfrac%7B3-11%7D%7B4%7D%3D-2%3C0-neyd%5C%5C%5C%5C%28%5Cfrac%7B7%7D%7B2%7D%29%5E%7B%5Csqrt%7Bx%2B1%7D%7D%3D%5Cfrac%7B7%7D%7B2%7D)
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