Chapter EP, Problem 10.2.6EP

To determine

To calculate; the vertex, the axis of symmetry and the direction of opening of parabola and write the equation in standard form, also graph the equation.

Answer to Problem 10.2.6EP

Solution:

  $x=2{(y-2)}^2-1$

Vertex $(h,k)=(-1,2)$ .

Axis of symmetry $(y=k)=y=2$ .

Direction of opening is right since $a>0$ .

Explanation of Solution

Write the equation in the form of $y=a{(x-h)}^2+k$ or $x=a{(y-k)}^2+h$ .

Given:- $x=2y^2-8y+7$

  $\begin{array}{l}x=2(y^2-4y+\frac{7}{2})\\ x=2(y^2-4y+4-4+\frac{7}{2})\\ x=2\left({(y-2)}^2-\frac{1}{2}\right)\\ $ x=2{(y-2)}^2-1$\end{array}$

Vertex $(h,k)=(-1,2)$ .

Axis of symmetry $(y=k)=y=2$ .

Direction of opening is right since $a>0$ .