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Solving Linear-Quadratic Systems

Solving Linear-Quadratic Systems 1.4 - Solution

arrow_back Return to 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,{\color{#0000FF}{a}}, b,{\color{#FF0000}{b}}, and c.{\color{#009600}{c}}. y=-x2+x1y=-1x2+1x+(-1)\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 a=-1,{\color{#0000FF}{a}}={\color{#0000FF}{\text{-} 1}}, b=1,{\color{#FF0000}{b}}={\color{#FF0000}{1}}, and c=-1.{\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 yy as a function of x,x, y=f(x).y=f(x). Vertex of a parabola: (-b2a,f(-b2a))\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 xx-coordinate of the vertex.
-b2a\text{-} \dfrac{b}{2a}
-12(-1)\text{-} \dfrac{{\color{#FF0000}{1}}}{2({\color{#0000FF}{\text{-} 1}})}
-1-2\text{-}\dfrac{1}{\text{-} 2}
We use the xx-coordinate of the vertex to find its yy-coordinate by substituting it into the given equation.
y=-x2+x1y=\text{-} x^2+x-1
y=-0.52+0.51y=\text{-} {\color{#0000FF}{0.5}}^2+{\color{#0000FF}{0.5}}-1
Simplify right-hand side
y=-0.25+0.51y=\text{-} 0.25+0.5-1
y=-0.75y=\text{-} 0.75
The yy-coordinate of the vertex is -0.75.\text{-} 0.75. Thus, the vertex is at the point (0.5,-0.75).(0.5,\text{-} 0.75). With this, we also know that the axis of symmetry of the parabola is the line x=0.5.x=0.5. Next, let's find two more points on the curve, one on each side of the axis of symmetry.
xx -x2+x1\text{-} x^2+x-1 y=-x2+x1y=\text{-} x^2+x-1
-1\textcolor{darkorange}{\text{-} 1} -(-1)2+(-1)1\text{-} (\textcolor{darkorange}{\text{-} 1})^2+(\textcolor{darkorange}{\text{-} 1})-1 -3\textcolor{purple}{\text{-} 3}
2\textcolor{darkorange}{2} -22+21\text{-} \textcolor{darkorange}{2}^2+\textcolor{darkorange}{2}-1 -3\textcolor{purple}{\text{-} 3}

Both (-1,-3)(\textcolor{darkorange}{\text{-} 1},\textcolor{purple}{\text{-} 3}) and (2,-3)(\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 m\textcolor{#ff8c00}{m} and yy-intercept b.{\color{#FF0000}{b}}. y=x1y=1x+(-1)\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 11 and the yy-intercept is -1.\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,-1).(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!
{y=x1(I)y=-x2+x1(II)\begin{cases}y=x-1 & \, \text {(I)}\\ y=\text{-} x^2+x-1 & \text {(II)}\end{cases}
(I), (II): {\color{#8C8C8C}{\text{(I), (II): }}} x=0x={\color{#0000FF}{0}}, y=-1y={\color{#009600}{\text{-} 1}}
{-1=?01-1=?-02+01\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
{-1=?01-1=?-0+01\begin{cases}\text{-} 1\stackrel{?}{=}0-1 \\ \text{-} 1\stackrel{?}{=}\text{-} 0+0-1 \end{cases}
(I), (II): {\color{#8C8C8C}{\text{(I), (II): }}} Add and subtract terms
{-1=-1 -1=-1 \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,-1)(0,\text{-} 1) is a correct solution.