Chapter 9.2, Problem 1GP

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 1GP

The standard equation is $x=-3{\left(y+2\right)}^2+30$ , vertex is $\left(30,-2\right)$ , axis of symmetry is $y=2$ and parabola opens towards left.

Explanation of Solution

Given:

The given equation is $y=4x^2+16x+34$ .

Calculation:

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

  $\begin{array}{c}y=4x^2+16x+34\\ =4\left(x^2+4x+4\right)-16+34\\ =4{\left(x+2\right)}^2+18\end{array}$

Therefore, the standard form of the given quadratic equation is $y=4{\left(x+2\right)}^2+18$ .

Compare the equation $y=4{\left(x+2\right)}^2+18$ with the standard equation $y=a{\left(x-h\right)}^2+k$ .

  $\begin{array}{c}a=4\\ h=-2\\ k=18\end{array}$

Therefore, the vertex of the given equation is $\left(-2,18\right)$ .

Axis of symmetry will be given shown below.

  $\begin{array}{c}y=k\\ =18\end{array}$

Therefore, the axis of symmetry is $y=18$ .

As it is found that the value of $a=4$ is positive so so by definition the parabola opens upward.

Therefore, the standard equation is $y=4{\left(x+2\right)}^2+18$ , vertex is $\left(-2,18\right)$ , axis of symmetry is $y=18$ and parabola opens upward.