91Ó°ÊÓ

We have:

$y=3x^2+6x+8$

The equation of a vertical parabola with vertex $\left(h,k\right)$and focus $\left(h,k+\frac{1}{4a}\right)$is given by:

$y=a{\left(x-h\right)}^2+k$

Convert the given equation into standard form:

$$\begin{gathered} y=3\left(x^2+2x\right)+8 \\ y+3=3\left(x^2+2x+1\right)+8 \\ y+3=3{\left(x+1\right)}^2+8 \\ y=3{\left(x+1\right)}^2+8-3 \\ y=3{\left(x+1\right)}^2+5 \end{gathered}$$

Compare the equation $y=3{\left(x+1\right)}^2+5$with standard form, we get:

$a=3$

A vertical parabola in standard form with a positive $a$ opens upward.

Therefore, the parabola opens upward.

The vertex of a vertical parabola in standard form $y=3{\left(x+1\right)}^2+5$is $\left(-1,5\right)$.

The axis of symmetry of a vertical parabola is given by:

$$\begin{gathered} x=-h \\ x=-1 \end{gathered}$$

We have:

$y=3x^2+6x+8$

To find x-intercept: Substitute $y=0:$

$3x^2+6x+8=0$

For a quadratic equation of the form $a{x}^2+bx+c=0$, the solutions are:

$x_{1, 2}=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For} a=3, b=6, c=8$

$$\begin{gathered} x_{1, 2}=\frac{-6Â\pm \sqrt{6^2-4· 3· 8}}{2· 3} \\ {x}_1=\frac{-6+2\sqrt{15}i}{2· 3}, x_2=\frac{-6-2\sqrt{15}i}{2· 3} \end{gathered}$$

No real solutions.

Hence, there is no x-intercept.

Now, find y- intercept by substituting $x=0:$

$$\begin{gathered} y=3{\left(0\right)}^2+6\left(0\right)+8 \\ y=8 \end{gathered}$$

$\mathrm{Y} \mathrm{Intercepts}: \left(0, 8\right)$

We have:

$y=3x^2+6x+8$

By comparing the equation with quadratic equation $a{x}^2+bx+c=0$, we get:

$a=3$

If $a<0$then the vertex is a minimum value.

If $a>0$then the vertex is a maximum value.

Here, $a=3>0$.

Hence, $\mathrm{Minimum}\left(-1, 5\right)$