Chapter 9.2, Problem 1CYU

To determine

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

Answer to Problem 1CYU

The standard equation is $y=2{\left(x-6\right)}^2-32$ , vertex is $\left(6,-32\right)$ , axis of symmetry is $x=6$ and parabola opens upward.

Explanation of Solution

Given:

The given equation is $y=2x^2-24x+40$ .

Calculation:

Rewrite the given quadratic equation in the standard form by completing the square.

  $\begin{array}{c}y=2x^2-24x+40\\ =2\left(x^2-12x+36\right)-72+40\\ y=2{\left(x-6\right)}^2-32\end{array}$

Therefore, the standard form of the given quadratic equation is $y=2{\left(x-6\right)}^2-32$ .

Compare the equation $y=2{\left(x-6\right)}^2-32$ with the standard equation $y=a{\left(x-h\right)}^2+k$ .

  $\begin{array}{c}a=2\\ h=6\\ k=-32\end{array}$

Therefore, the vertex of the given equation is $\left(6,-32\right)$ .

Axis of symmetry will be given shown below.

  $\begin{array}{c}x=h\\ =6\end{array}$

Therefore, the axis of symmetry is $x=6$ .

As it is found that the value of $a=2$ is positive so by definition the parabola opens upward.

Therefore, the standard equation is $y=2{\left(x-6\right)}^2-32$ , vertex is $\left(6,-32\right)$ , axis of symmetry is $x=6$ and parabola opens upward.