To solve the by graphing, we will draw the graph of the two on the same coordinate grid. Let's start with the first equation.
Graphing the First Parabola
To graph the we first need to identify
$a,$ $b,$ and
$c.$
$y=x_{2}+8x+18⇔y=1x_{2}+8x+18 $
For this equation we have that
$a=1,$ $b=8,$ and
$c=18.$ Now, we can find the using its formula. To do this, we will need to think of
$y$ as a function of
$x,$ $y=f(x).$
$Vertex of a parabola:(2ab ,f(2ab )) $
Let's find the
$x$coordinate of the vertex.
$2ab $
$2(1)8 $
$28 $
$4$
We use the
$x$coordinate of the vertex to find its
$y$coordinate by substituting it into the given equation.
$y=x_{2}+8x+18$
$y=(4)_{2}+8(4)+18$
$y=16+8(4)+18$
$y=16−32+18$
$y=2$
The
$y$coordinate of the vertex is
$2.$ Thus, the vertex is at the point
$(4,2).$ With this, we also know that the of the parabola is the line
$x=4.$ Next, let's find two more points on the curve, one on each side of the axis of symmetry.
$x$

$x_{2}+8x+18$

$y=x_{2}+8x+18$

$5$

$(5)_{2}+8(5)+18$

$3$

$3$

$(3)_{2}+8(3)+18$

$3$

Both $(5,3)$ and $(3,3)$ are on the graph. Let's form the parabola by connecting these points and the vertex with a smooth curve.
Graphing the Second Parabola
Just like above, we will start with identifying
$a,$ $b,$ and
$c.$
$y=2x_{2}−16x−30⇕y=2x_{2}+(16)x+(30) $
For this equation we have that
$a=2,$ $b=16,$ and
$c=30.$ Let's again find the
$x$coordinate of the .
$2ab $
$2(2)16 $
$416 $
$4$
We use the
$x$coordinate of the vertex to find its
$y$coordinate by substituting it into the given equation.
$y=2x_{2}−16x−30$
$y=2(4)_{2}−16(4)−30$
$y=2(16)−16(4)−30$
$y=32+64−30$
$y=3$
The
$y$coordinate of the vertex is
$3.$ Thus, the vertex is at the same point as the previous parabola. The next step is to find two more points on the curve, one on each side of the axis of symmetry, which is
$x=4.$
$x$

$2_{2}−16x−30$

$y=2x_{2}−16x−30$

$5$

$2(5)_{2}−16(5)−30$

$0$

$3$

$2(3)_{2}−16(3)−30$

$0$

Both $(5,0)$ and $(3,0)$ are on the graph. Let's form the second parabola by connecting these points and the vertex with a smooth curve.
Finding the Solutions
Finally, let's try to identify the coordinates of the of the two parabolas.
It looks like the point of intersection occurs at $(4,2).$
Checking the Answer
To check our answer, we will substitute the values of the point of intersection in both equations of the system. If they produce true statements, our solution is correct.
${y=x_{2}+8x+18y=2x_{2}−16x−30 (I)(II) $
$(I), (II):$ $x=4$, $y=2$
${2=?(4)_{2}+8(4)+182=?2(4)_{2}−16(4)−30 $
$(I), (II):$ Calculate power
${2=?16+8(4)+182=?2(16)−16(4)−30 $
$(I), (II):$ Multiply
${2=?16−32+182=?32+64−30 $
$(I), (II):$ Add and subtract terms
${2=2✓2=2✓ $
Equation (I) and Equation (II) both produced true statements. Therefore,
$(4,2)$ is the correct solution.