Answer:
[tex]\textsf{a)} \quad y=(x+1)^2-4[/tex]
[tex]\textsf{b)} \quad y=2(x-4)^2-2[/tex]
Step-by-step explanation:
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)
Part a
- Vertex = (-1, -4)
- Point on parabola = (1, 0)
Substitute the vertex into the formula:
[tex]\begin{aligned}\implies y &=a(x-(-1))^2-4\\y & =a(x+1)^2-4\end{aligned}[/tex]
Substitute the point (1, 0) into the formula:
[tex]\begin{aligned}\implies a(1+1)^2-4&=0\\4a-4&=0\\4a &=4\\a &=1\end{aligned}[/tex]
Therefore, the equation of the parabola in vertex form is:
[tex]y=(x+1)^2-4[/tex]
Part b
- Vertex = (4, -2)
- Point on parabola = (6, 6)
Substitute the vertex into the formula:
[tex]\begin{aligned}\implies y &=a(x-4)^2+(-2)\\y & =a(x-4)^2-2\end{aligned}[/tex]
Substitute the point (6, 6) into the formula:
[tex]\begin{aligned}\implies a(6-4)^2-2&=6\\4a-2 & =6\\4a &=8\\a &=2\end{aligned}[/tex]
Therefore, the equation of the parabola in vertex form is:
[tex]y=2(x-4)^2-2[/tex]
