Answer:
[tex]\left(-\dfrac{1}{3},-\dfrac{4}{3}\right)[/tex]
Step-by-step explanation:
The graph of the parabola passes through the points (-1,0), (0,-1) and (1,4).
Let [tex]y=ax^2+bx+c[/tex] be the equation of the parabola. Substitute the coordinates of the points:
[tex]0=a\cdot (-1)^2+b\cdot (-1)+c\Rightarrow a-b+c=0\\ \\-1=a\cdot 0^2+b\cdot 0+c\Rightarrow c=-1\\ \\4=a\cdot 1^2+b\cdot 1+c\Rightarrow a+b+c=4[/tex]
Substitute [tex]c=-1[/tex] into the first and third equations:
[tex]a-b-1=0\Rightarrow a-b=1\\ \\a+b-1=4\Rightarrow a+b=5[/tex]
Add them:
[tex]a-b+a+b=1+5\\ \\2a=6\\ \\a=3\\ \\3-b=1\\ \\b=3-1=2[/tex]
Hence,
[tex]y=3x^2+2x-1[/tex]
Find the vertex:
[tex]x_v=\dfrac{-b}{2a}=\dfrac{-2}{2\cdot 3}=-\dfrac{1}{3}\\ \\y_v=3\cdot \left(-\dfrac{1}{3}\right)^2+2\cdot \left(-\dfrac{1}{3}\right)-1=\dfrac{1}{3}-\dfrac{2}{3}-1=-\dfrac{4}{3}[/tex]
So, the vertex has coordinates
[tex]\left(-\dfrac{1}{3},-\dfrac{4}{3}\right)[/tex]
