Solving Linear-Quadratic Systems

We will graph the system of equations on one coordinate grid. Using the graph we will then solve the system. Let's start by graphing the parabola.

Graphing the Parabola

To graph the parabola, we first need to identify $a,$ $b,$ and $c.$ $$\begin{array}{c}y=x^2+2x+1.5\iff y=1x^2+2x+1.5\end{array}$$ For this equation we have that $a=1,$ $b=2,$ and $c=1.5.$ 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 use the $x$-coordinate of the vertex to find its $y$-coordinate by substituting it into the given equation. The $y$-coordinate of the vertex is $\text{-}0.5.$ Thus, the vertex is at the point $(\text{-}1,\text{-}0.5).$ With this, we also know that the axis of symmetry of the parabola is the line $x=\text{-}1.$ Next, let's find two more points on the curve, one on each side of the axis of symmetry.

$x$	$x^2+2x+1.5$	$y$
$\text{-}1.5$	$(\text{-}1.5{)}^2+2(\text{-}1.5)+1.5$	$0.75$
$\text{-}0.5$	$(\text{-}0.5{)}^2+2(\text{-}0.5)+1.5$	$0.75$

Both $(\text{-}1.5,0.75)$ and $(\text{-}0.5,0.75)$ 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 $b.$ $$\begin{array}{c}y=\text{-}x-1\iff y=\text{-}1x+(\text{-}1)\end{array}$$ The slope of the line is $\text{-}1$ and the $y$-intercept is $\text{-}3.$

We can see that our graph corresponds to option D.

Finding the Solutions

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

We can see that the graphs do not intersect. Therefore, the given system of equations has no solutions.