Determine the vertex of the following parabola:
y=2\left(x-1\right)^2+1
When a parabola is given in the vertex form:
y=a\left(x-h\right)^2+k
The vertex of a parabola is \left(h,k\right).
Here, we have:
y=2\left(x-1\right)^2+1
The vertex of the parabola is (1,1).
Determine the focus of the following parabola:
y=\left(x-2\right)^2+3
When a parabola is given in the vertex form:
y=a\left(x-h\right)^2+k
The vertex of a parabola is \left(h,k\right) and the focus is \left(h,k+\dfrac{1}{4a}\right).
Here, we have:
- \left(h,k\right)=\left(2{,}3\right)
- F=\left(2{,}3+\dfrac{1}{4}\right)=\left(2,\dfrac{13}{4}\right)
The focus of the parabola is \left(2,\dfrac{13}{4}\right).
Determine the focus of the following parabola:
x=\dfrac{1}{2}\left(y+1\right)^2+2
When a parabola is given in the vertex form:
x=a\left(y-k\right)^2+h
The vertex of a parabola is \left(h,k\right) and the focus is \left(h+\dfrac{1}{4a},k\right).
Here, we have:
- \left(h,k\right)=\left(2,-1\right)
- F=\left(2+\dfrac{1}{\left(4\times \dfrac{1}{2}\right)},-1\right)=\left(2+\dfrac{1}{2},-1\right)=\left(\dfrac{5}{2},-1\right)
The focus of the parabola is \left(\dfrac{5}{2},-1\right).
Determine the vertex of the following parabola:
x=y^2+6y+10
When a parabola is given in the vertex form:
x=a\left(y-k\right)^2+h
The vertex of a parabola is \left(h,k\right). Find the vertex form of the parabola:
x=y^2+6y+10
x=\left(y^2+6y+9\right)+1
x=\left(y+3\right)^2+1
Therefore:
The vertex of the parabola is (1,-3).
Determine the vertex of the following parabola:
y=2x^2+16x+40
When a parabola is given in the vertex form:
y=a\left(x-h\right)^2+k
The vertex of a parabola is \left(h,k\right). Find the vertex form of the parabola:
Here we have:
y=2x^2+16x+40
y=2\left(x^2+8x+ 16\right)+8
y=2\left(x+4\right)^2+8
The vertex of the parabola is (-4,8).
Determine the focus of the following parabola:
y=x^2+4x+5
When a parabola is given in the vertex form:
y=a\left(x-h\right)^2+k
The vertex of a parabola is \left(h,k\right) and the focus is \left(h,k+\dfrac{1}{4a}\right). Find the vertex form of the parabola:
y=x^2+4x+5
y=\left(x^2+4x+4\right)+1
y=\left(x+2\right)^2+1
So we have:
- \left(h,k\right)=\left(-2{,}1\right)
- F=\left(-2{,}1+\dfrac{1}{4}\right)=\left(-2,\dfrac{5}{4}\right)
The focus of the parabola is \left(-2,\dfrac{5}{4}\right).
Determine the focus of the following parabola:
x=y^2+6y+12
When a parabola is given in the vertex form:
x=a\left(y-k\right)^2+h
The vertex of a parabola is \left(h,k\right) and the focus is \left(h+\dfrac{1}{4a},k\right). Find the vertex form of the parabola:
x=y^2+9x+12
x=\left(y^2+6y+9\right)+3
x=\left(y+3\right)^2+3
So we have:
- \left(h,k\right)=\left(3,-3\right)
- F=\left(3+\dfrac{1}{4},-3\right)=\left(\dfrac{13}{4},-3\right)
The focus of the parabola is \left(\dfrac{13}{4},-3\right).