Chapter 6.4, Problem 86SR

To determine

Write the equation $y=2x^2+8x+10$ in vertex form and then identify vertex, axis of symmetry, direction of opening and sketch the graph of the equation.

Answer to Problem 86SR

The vertex is $\left(-2,2\right)$ , axis of symmetry is $x=-2$ and direction of opening of graph is upwards.

Explanation of Solution

Given information: The given equation is,

  $y=2x^2+8x+\mathrm{10......}\left(1\right)$

Calculation: Changing the equation (1) to vertex form as,

  $\begin{array}{c}y=2x^2+8x+\mathrm{10......}\left(\text{Given}\right)\\ y=2\left(x^2+4x+5\right)\mathrm{......}\left(\text{Rearranged}\right)\\ y=2\left({\left(x\right)}^2+\left(2\right)\left(x\right)\left(2\right)+{\left(2\right)}^2-{\left(2\right)}^2+5\right)\\ y=2\left({\left(x+2\right)}^2+1\right)\\ y=2{\left(x+2\right)}^2+\mathrm{2......}\left(\text{Vertex-form}\right)\end{array}$

Comparing with vertex form $y=a{\left(x-h\right)}^2+k$ of a standard parabola, the vertex is $\left(-2,2\right)$ .

Now the vertex is the point $\left(-2,2\right)$ , therefore the equation for axis of symmetry is the line passing through the $\left(x\right)$ coordinate of the vertex. Therefore the equation for axis of symmetry is $x=-2$ .

From the vertex form of parabola the leading coefficient of ${\left(x+2\right)}^2$ is 2 which is positive therefore the parabola will be upward parabola which will open upwards.

Graph: The graph for the function is shown as,