We will first express the equation in standard form, x=a(y-k)^2+h, where a, h, and k are either positive or negative constants. To do so, we will simplify the right-hand side of the equation by completing the square. We have to add and subtract ( b2)^2. In this case, we have that the linearcoefficient b is 14.
b=14 ⇒ (b/2)^2=(14/2)^2=7^2
Let's do it!
In standard form the equation is written as x=a(y-k)^2+h, where a, h, and k are either positive or negative constants.
x=(y+7)^2-29
⇕
x=1(y-(- 7))^2+(-29)
It is important to note that we do not need to graph the parabola to identify the desired information. Let's compare the general formula for the standard form to our equation.
General Formula:x=& a(y- k )^2 + h
Equation:x=& 1(y-( - 7))^2+(-29)
We can see that a= 1, k= - 7, and h=- 29.
Vertex
The vertex of a quadratic function written in standard form is the point (h, k). For this exercise, we have h=- 29 and k= - 7 . Therefore, the vertex of the given equation is (- 29, - 7).
Axis of Symmetry
The axis of symmetry of the quadratic quadratic equation written in standard form is the horizontal line with equation y= k. As we have already noticed, for our equation, this is k= - 7. Thus, the axis of symmetry is the line y= -7.
Direction of Opening
Recall that when the axis of symmetry is a horizontal line, if a>0, the parabola opens right. Conversely, if a<0, the parabola opens left. In the given equation, we have a= 1, which is greater than 0. Thus, the parabola opens right.
