We will graph the on one coordinate grid. Using the graph we will then solve the system. Let's start by graphing the .
Graphing the Parabola
To , we first need to identify
$a,$ $b,$ and
$c.$
$y=x_{2}−2x+1⇔y=1⋅x_{2}+(2)x+1 $
For this equation we have that
$a=1,$ $b=2,$ and
$c=1.$ 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⋅12 $
$22 $
$22 $
$1$
We use the
$x$coordinate of the vertex to find its
$y$coordinate by substituting it into the given equation.
$y=x_{2}−2x+1$
$y=1_{2}−2⋅1+1$
$y=0$
The
$y$coordinate of the vertex is
$0.$ Thus, the vertex is at the point
$(1,0).$ 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.
$x$

$x_{2}−2x+1$

$y=x_{2}−2x+1$

$1$

$(1)_{2}−2(1)+1$

$4$

$3$

$3_{2}−2(3)+1$

$4$

Both $(1,4)$ and $(3,4)$ 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=x+1⇔y=1x+1 $
The slope of the line is $1$ and the $y$intercept is $1.$
We can see that our graph corresponds to option B.
Finding the Solutions
Finally, let's try to identify the coordinates of the of the parabola and the line.
It looks like the points of intersection occur at $(0,1)$ and $(3,4).$
Checking the Answer
To check our answers, we will substitute the values of the points of intersection in both equations of the system. If they produce true statements, our solution is correct. Let's start with
$(0,1).$
${y=x_{2}−2x+1y=x+1 (I)(II) $
$(I), (II):$ $x=0$, $y=1$
${1=?0_{2}−2(0)+11=?0+1 $
${1=?0−2(0)+11=?0+1 $
${1=?0−0+11=?0+1 $
$(I), (II):$ Add and subtract terms
${1=1✓1=1✓ $
Equation (I) and Equation (II) both produced true statements. Therefore,
$(0,1)$ is a correct solution. Let's continue by checking
$(3,4).$
${y=x_{2}−2x+1y=x+1 (I)(II) $
$(I), (II):$ $x=3$, $y=4$
${4=?3_{2}−2(3)+14=?3+1 $
${4=?9−2(3)+14=?3+1 $
${4=?9−6+14=?3+1 $
$(I), (II):$ Add and subtract terms
${4=4✓4=4✓ $
Equation (I) and Equation (II) produced true statements again. Therefore,
$(3,4)$ is also a correct solution.