Chapter 9.4, Problem 6MCQ

To determine

To find: The given equation in the standard form and the vertex, axis of symmetry, and the direction of opening of the parabola.

Answer to Problem 6MCQ

The given equation in the standard form is $y=3{\left(x-2\right)}^2+9$ , the coordinate of the vertex is $\left(2,9\right)$ , the axis of symmetry is $x=2$ , and the direction of opening of the parabola is upward.

Explanation of Solution

Given information:

The equation of the parabola is,

  $y=3x^2-12x+21$

Concept used:

Write the standard form of the equation of the parabola,

  $y=a{\left(x-h\right)}^2+k$

The direction of opening,

  $\begin{array}{l}\text{upward, }a>0\\ \text{downward, }a<0\end{array}$

The coordinate of the vertex is $\left(h,k\right)$ , the coordinate of focus is $\left(h,k+\frac{1}{4a}\right)$ , and the length of latus rectum is $\left|\frac{1}{a}\right|$ .

The equation of directrix is,

  $y=k-\frac{1}{4a}$

The axis of symmetry is,

  $x=h$

Write the standard form of the equation of the parabola,

  $\begin{array}{l}y=3x^2-12x+21\\ y=3\left(x^2-4x+4\right)-12+21\\ y=3{\left(x-2\right)}^2+9\end{array}$

The direction of opening is,

  $a=3>0,\text{ upward}$

The coordinate of the vertex is $\left(2,9\right)$ .

The axis of symmetry is,

  $\begin{array}{l}x=h\\ x=2\end{array}$

Thus, the given equation in the standard form is $y=3{\left(x-2\right)}^2+9$ , the coordinate of the vertex is $\left(2,9\right)$ , the axis of symmetry is $x=2$ , and the direction of opening of the parabola is upward.