Respuesta :
Answer: y = 2 (x minus one-half) squared minus StartFraction 27 Over 2 EndFraction
or
[tex]y=2((x-\dfrac{1}{2})^2)-\dfrac{27}{2}[/tex]
Step-by-step explanation:
Vertex form of equation : [tex]f (x) = a(x - h)^2 + k,[/tex]where (h, k) is the vertex of the parabola.
[tex]y=3(x-2)^2-(x-5)^2\\\\=3(x^2+4-4x)-(x^2+25-10x)\\\\=3x^2+12-12x-x^2-25+10x\\\\=2x^2-2x-13\\\\=2(x^2-x-\dfrac{13}{2})\\\\=2(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{13}{2})\\\\=2((x-\dfrac{1}{2})^2-\dfrac{1+26}{4})\\\\=2((x-\dfrac{1}{2})^2-\dfrac{27}{4})=2((x-\dfrac{1}{2})^2)-\dfrac{27}{2}[/tex]
Hence, the vertex form of the equation is [tex]y=2((x-\dfrac{1}{2})^2)-\dfrac{27}{2}[/tex]
Answer:
i got y=2(x-1/2)^2- 27/2 so the last one is right
Step-by-step explanation:
:)