Answer:
vertex = (0, -4)
equation of the parabola: [tex]y=3x^2-4[/tex]
Step-by-step explanation:
Given:
- y-intercept of parabola: -4
- parabola passes through points: (-2, 8) and (1, -1)
Vertex form of a parabola: [tex]y=a(x-h)^2+k[/tex]
(where (h, k) is the vertex and [tex]a[/tex] is some constant)
Substitute point (0, -4) into the equation:
[tex]\begin{aligned}\textsf{At}\:(0,-4) \implies a(0-h)^2+k &=-4\\ah^2+k &=-4\end{aligned}[/tex]
Substitute point (-2, 8) and [tex]ah^2+k=-4[/tex] into the equation:
[tex]\begin{aligned}\textsf{At}\:(-2,8) \implies a(-2-h)^2+k &=8\\a(4+4h+h^2)+k &=8\\4a+4ah+ah^2+k &=8\\\implies 4a+4ah-4&=8\\4a(1+h)&=12\\a(1+h)&=3\end{aligned}[/tex]
Substitute point (1, -1) and [tex]ah^2+k=-4[/tex] into the equation:
[tex]\begin{aligned}\textsf{At}\:(1.-1) \implies a(1-h)^2+k &=-1\\a(1-2h+h^2)+k &=-1\\a-2ah+ah^2+k &=-1\\\implies a-2ah-4&=-1\\a(1-2h)&=3\end{aligned}[/tex]
Equate to find h:
[tex]\begin{aligned}\implies a(1+h) &=a(1-2h)\\1+h &=1-2h\\3h &=0\\h &=0\end{aligned}[/tex]
Substitute found value of h into one of the equations to find a:
[tex]\begin{aligned}\implies a(1+0) &=3\\a &=3\end{aligned}[/tex]
Substitute found values of h and a to find k:
[tex]\begin{aligned}\implies ah^2+k&=-4\\(3)(0)^2+k &=-4\\k &=-4\end{aligned}[/tex]
Therefore, the equation of the parabola in vertex form is:
[tex]\implies y=3(x-0)^2-4=3x^2-4[/tex]
So the vertex of the parabola is (0, -4)
