To solve the by graphing, we will draw the graph of the and the on the same . Let's start with the .
Graphing the Parabola
To graph the parabola, we first need to identify
$a,$ $b,$ and
$c.$
$y=x_{2}+2x+5⇔y=1x_{2}+2x+5 $
For this equation we have that
$a=1,$ $b=2,$ and
$c=5.$ 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)2 $
$22 $
$1$
We will use the
$x$coordinate of the vertex to find its
$y$coordinate by substituting it into the given equation.
$y=x_{2}+2x+5$
$y=(1)_{2}+2(1)+5$
$y=4$
The
$y$coordinate of the vertex is
$4.$ Thus, the vertex is the point
$(1,4).$ With this, we also know that the of the parabola is the line
$x=1.$ Next, let's find two more points on the curve, one on each side of the axis of symmetry.
$x$

$x_{2}+2x+5$

$y=x_{2}+2x+5$

$0$

$0_{2}+2(0)+5$

$5$

$2$

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

$5$

Both $(0,5)$ and $(2,5)$ are on the graph. Let's form the parabola by connecting these points and the vertex with a smooth curve.
Graphing the Line
Let's now graph the linear function on the same coordinate plane. For a linear equation written in , we can identify its slope $m$ and $y$intercept $b.$
$y=2x+1 $
The slope of the line is $2$ and the $y$intercept is $1.$
Finding the Solutions
Finally, let's try to identify the coordinates of the of the parabola and the line.
It looks like the only point of intersection occurs at
$(2,5).$ To check our answer, we will substitute the coordinates of the point of intersection in both equations of the system. If they produce true statements, our solution is correct.
${y=x_{2}+2x+5y=2x+1 (I)(II) $
$(I), (II):$ $x=2$, $y=5$
${5=?(2)_{2}+2(2)+55=?2(2)+1 $
${5=?4+2(2)+55=?2(2)+1 $
${5=?4−4+55=?2(2)+1 $
${5=?4−4+55=?4+1 $
$(I), (II):$ Add and subtract terms
${5=5✓5=5✓ $
Equation (I) and Equation (II) both produced true statements. Therefore,
$(2,5)$ is a correct solution.