Solving Linear-Quadratic Systems

To solve the system of equations by graphing, we will draw the graph of the quadratic function and the linear function on the same coordinate system. Let's start with the parabola.

Graphing the Parabola

To graph the parabola, we first need to identify $a,$ $b,$ and $c.$ $$\begin{array}{c}y=x^2+1\iff y=1x^2+0x+1\end{array}$$ For this equation we have that $a=1,$ $b=0,$ and $c=1.$ Now, we can find the vertex using its formula. To do this, we will need to think of $y$ as a function of $x,$ $y=f(x).$ $$\begin{array}{c}\text{Vertex of a parabola:}\left(\text{-}\frac{b}{2a},f\left(\text{-}\frac{b}{2a}\right)\right)\end{array}$$ Let's find the $x$-coordinate of the vertex. We will use the $x\text{-}$coordinate of the vertex to find its $y\text{-}$coordinate by substituting it into the given equation. The $y\text{-}$coordinate of the vertex is $1.$ Therefore, the vertex is the point $(0,1).$ With this, we also know that the axis of symmetry of the parabola is the line $x=0.$ Next, let's find two more points on the curve, one on each side of the axis of symmetry.

$x$	$x^2+1$	$y=x^2+1$
$\text{-}1$	$(\text{-}1{)}^2+1$	$2$
$1$	$1^2+1$	$2$

Both $(\text{-}1,2)$ and $(1,2)$ 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 slope-intercept form, we can identify its slope $m$ and $y$-intercept $b.$ $$\begin{array}{c}y=x+1\iff y=1x+1\end{array}$$ The slope of the line is $1$ and the $y\text{-}$intercept is $1.$

Finding the Solutions

Finally, let's try to identify the coordinates of the points of intersection of the parabola and the line.

It looks like the points of intersection occur at $(0,1)$ and $(1,2).$