Question

Graph the following parabolas by using intercepts, the vertex, and the axis of symmetry.

$y=2x^2+6x+2$

Short answer

The required graph is shown below:

Step-by-step solution

Given information.

The given equation is$y=2x^2+6x+2$

Determine the axis of symmetry.

By comparing the given equation with the standard form $y=a{x}^2+bx+c$we get$a=2,b=6andc=2$

Since the value of a is positive so parabola opens up.

The axis of symmetry is the vertical line:

$$\begin{gathered} x=-\frac{b}{2a} \\ =-\frac{6}{2\left(2\right)} \\ =-\frac{3}{2} \end{gathered}$$

The axis of symmetry is $x=-\frac{3}{2}$.

Determine the vertex.

Put $x=-\frac{3}{2}$into the given equation we get.

$$\begin{gathered} y=2{\left(-\frac{3}{2}\right)}^2+6\left(-\frac{3}{2}\right)+2 \\ =\frac{9}{2}-9+2 \\ =-\frac{5}{2} \end{gathered}$$

The vertex is$\left(-\frac{3}{2},-\frac{5}{2}\right)$.

Determine the intercepts.

For y-intercept put $x=0$into the given equation.

$$\begin{gathered} y=2{\left(0\right)}^2+6\left(0\right)+2 \\ =2 \end{gathered}$$

The y-intercept is $(0,2)$.

For x-intercept put $y=0$into the given equation.

$$\begin{gathered} 2x^2+6x+2=0 \\ x=\frac{-6\pm \sqrt{6^2-4\left(2\right)\left(2\right)}}{2\left(2\right)} \\ =\frac{-3\pm \sqrt{5}}{2} \end{gathered}$$

The x-intercept is role="math" localid="1653824362529" $\left(\frac{-3+\sqrt{5}}{2},0\right)and\left(\frac{-3-\sqrt{5}}{2},0\right)$.

Draw the graph.

By using the above information the required graph is shown below: