Find the equations of the following parabolas.
One equation of a vertical parabola is:
y=k\left(x-a\right)^2+ b
Where \left(a,b\right) is the vertex of the parabola.
According to the graph, the vertex is \left(1{,}1\right). So we have:
y=k\left(x-1\right)^2+ 1
To find k, we choose an arbitrary point \left(x_0,y_0\right) on the parabola and substitute x_0 for x and y_0 for y. Here, we can choose \left(0{,}2\right) :
2=k\left(0-1\right)^2+ 1
2=k+ 1
k=1
Therefore, an equation of the parabola is:
y=\left(x-1\right)^2+1
One equation of a vertical parabola is:
y=k\left(x-a\right)^2+ b
Where \left(a,b\right) is the vertex of the parabola.
According to the graph, the vertex is \left(-1{,}1\right). So we have:
y=k\left(x+1\right)^2+ 1
To find k, we choose an arbitrary point \left(x_0,y_0\right) on the parabola and substitute x_0 for x and y_0 for y. Here, we can choose \left(0,-1\right) :
-1=k\left(0+1\right)^2+ 1
-2=k
k=-2
Therefore, an equation of the parabola is:
y=-2\left(x+1\right)^2+1
One equation of a horizontal parabola is:
x=k\left(y-b\right)^2+ a
Where \left(a,b\right) is the vertex of the parabola.
According to the graph, the vertex is \left(1,-1\right). So we have:
x=k\left(y+1\right)^2+ 1
To find k, we choose an arbitrary point \left(x_0,y_0\right) on the parabola and substitute x_0 for x and y_0 for y. Here, we can choose \left(2{,}0\right) :
2=k\left(0+1\right)^2+ 1
2=k+ 1
k=1
Therefore, an equation of the parabola is:
x=\left(y+1\right)^2+1
One equation of a horizontal parabola is:
x=k\left(y-b\right)^2+ a
Where \left(a,b\right) is the vertex of the parabola.
According to the graph, the vertex is \left(3,-2\right). So we have:
x=k\left(y+2\right)^2+3
To find k, we choose an arbitrary point \left(x_0,y_0\right) on the parabola and substitute x_0 for x and y_0 for y. Here, we can choose \left(2,-1\right) :
2=k\left(-1+2\right)^2+ 3
2=k+ 3
k=-1
Therefore, an equation of the parabola is:
x=-\left(y+2\right)^2+3
One equation of a horizontal parabola is:
x=k\left(y-b\right)^2+ a
Where \left(a,b\right) is the vertex of the parabola.
According to the graph, the vertex is \left(-2{,}3\right). So we have:
x=k\left(y-3\right)^2-2
To find k, we choose an arbitrary point \left(x_0,y_0\right) on the parabola and substitute x_0 for x and y_0 for y. Here, we can choose \left(-1{,}2\right) :
-1=k\left(2-3\right)^2-2
-1=k-2
k=1
Therefore, an equation of the parabola is:
x=\left(y-3\right)^2-2
One equation of a vertical parabola is:
y=k\left(x-a\right)^2+ b
Where \left(a,b\right) is the vertex of the parabola.
According to the graph, the vertex is \left(-3{,}3\right). So we have:
y=k\left(x+3\right)^2+ 3
To find k, we choose an arbitrary point \left(x_0,y_0\right) on the parabola and substitute x_0 for x and y_0 for y. Here, we can choose \left(-1{,}7\right) :
7=k\left(-1+3\right)^2+ 3
7=4k+ 3
4k=4
k=1
Therefore, an equation of the parabola is:
y=\left(x+3\right)^2+3
One equation of a vertical parabola is:
y=k\left(x-a\right)^2+ b
Where \left(a,b\right) is the vertex of the parabola.
According to the graph, the vertex is \left(-1,-2\right). So we have:
y=k\left(x+1\right)^2-2
To find k, we choose an arbitrary point \left(x_0,y_0\right) on the parabola and substitute x_0 for x and y_0 for y. Here, we can choose \left(1,-4\right) :
-4=k\left(1+1\right)^2-2
-2=4k
k=-\dfrac{1}{2}
Therefore, an equation of the parabola is:
y=-\dfrac{1}{2}\left(x+1\right)^2-2