Expand menu menu_open Minimize Start chapters Home History history History expand_more
{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics equalizer Progress expand_more
Student
navigate_next
Teacher
navigate_next
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
No results
{{ searchError }}
search
{{ courseTrack.displayTitle }}
{{ statistics.percent }}% Sign in to view progress
{{ printedBook.courseTrack.name }} {{ printedBook.name }}

# Solving Linear-Quadratic Systems

## Solving Linear-Quadratic Systems 1.4 - Solution

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 ${\color{#0000FF}{a}},$ ${\color{#FF0000}{b}},$ and ${\color{#009600}{c}}.$ $\begin{gathered} y=\text{-} x^2+x-1 \quad \Leftrightarrow \quad y={\color{#0000FF}{\text{-} 1}}x^2+{\color{#FF0000}{1}}x+({\color{#009600}{\text{-} 1}}) \end{gathered}$ For this equation we have that ${\color{#0000FF}{a}}={\color{#0000FF}{\text{-} 1}},$ ${\color{#FF0000}{b}}={\color{#FF0000}{1}},$ and ${\color{#009600}{c}}={\color{#009600}{\text{-} 1}}.$ Now, we can find the vertex using its formula. To do this, we will need to think of $y$ as a function of $x,$ $y=f(x).$ $\begin{gathered} \textbf{Vertex of a parabola:}\ \left( \text{-} \dfrac{{\color{#FF0000}{b}}}{2{\color{#0000FF}{a}}},f\left(\text{-} \dfrac{{\color{#FF0000}{b}}}{2{\color{#0000FF}{a}}}\right) \right) \end{gathered}$ Let's find the $x$-coordinate of the vertex.
$\text{-} \dfrac{b}{2a}$
$\text{-} \dfrac{{\color{#FF0000}{1}}}{2({\color{#0000FF}{\text{-} 1}})}$
$\text{-}\dfrac{1}{\text{-} 2}$
$\dfrac{1}{2}$
$0.5$
We use the $x$-coordinate of the vertex to find its $y$-coordinate by substituting it into the given equation.
$y=\text{-} x^2+x-1$
$y=\text{-} {\color{#0000FF}{0.5}}^2+{\color{#0000FF}{0.5}}-1$
Simplify right-hand side
$y=\text{-} 0.25+0.5-1$
$y=\text{-} 0.75$
The $y$-coordinate of the vertex is $\text{-} 0.75.$ Thus, the vertex is at the point $(0.5,\text{-} 0.75).$ With this, we also know that the axis of symmetry of the parabola is the line $x=0.5.$ Next, let's find two more points on the curve, one on each side of the axis of symmetry.
$x$ $\text{-} x^2+x-1$ $y=\text{-} x^2+x-1$
$\textcolor{darkorange}{\text{-} 1}$ $\text{-} (\textcolor{darkorange}{\text{-} 1})^2+(\textcolor{darkorange}{\text{-} 1})-1$ $\textcolor{purple}{\text{-} 3}$
$\textcolor{darkorange}{2}$ $\text{-} \textcolor{darkorange}{2}^2+\textcolor{darkorange}{2}-1$ $\textcolor{purple}{\text{-} 3}$

Both $(\textcolor{darkorange}{\text{-} 1},\textcolor{purple}{\text{-} 3})$ and $(\textcolor{darkorange}{2},\textcolor{purple}{\text{-} 3})$ 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 $\textcolor{#ff8c00}{m}$ and $y$-intercept ${\color{#FF0000}{b}}.$ $\begin{gathered} y=x-1 \quad \Leftrightarrow \quad y=\textcolor{#ff8c00}{1}x+({\color{#FF0000}{\text{-} 1}}) \end{gathered}$ The slope of the line is $1$ and the $y$-intercept is $\text{-} 1.$

We can see that our graph corresponds to option A.

### 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 point of intersection occurs at $(0,\text{-} 1).$

### Checking the Answer

To check our answer, we will substitute the values of the point of intersection in both equations of the system. If it produces true statements, our solution is correct. Let's do it!
$\begin{cases}y=x-1 & \, \text {(I)}\\ y=\text{-} x^2+x-1 & \text {(II)}\end{cases}$
${\color{#8C8C8C}{\text{(I), (II): }}}$ $x={\color{#0000FF}{0}}$, $y={\color{#009600}{\text{-} 1}}$
$\begin{cases}{\color{#009600}{\text{-} 1}}\stackrel{?}{=}{\color{#0000FF}{0}}-1 \\ {\color{#009600}{\text{-} 1}}\stackrel{?}{=}\text{-}{\color{#0000FF}{0}}^2+{\color{#0000FF}{0}}-1 \end{cases}$
Simplify right-hand side
$\begin{cases}\text{-} 1\stackrel{?}{=}0-1 \\ \text{-} 1\stackrel{?}{=}\text{-} 0+0-1 \end{cases}$
${\color{#8C8C8C}{\text{(I), (II): }}}$ Add and subtract terms
$\begin{cases}\text{-} 1=\text{-} 1\ \checkmark \\ \text{-} 1=\text{-} 1\ \checkmark \end{cases}$
Equation (I) and Equation (II) both produced true statements. Therefore, $(0,\text{-} 1)$ is a correct solution.