Quadratic equations: Intersection points of parabolas
Intersection points of a parabola with a line
A quadratic formula #y=a_1x^2+b_1x+c_1# and a linear formula #y=a_2x+b_2# can have zero, one or two intersection points. We will now investigate how to find these intersection points.
Intersection points parabola and line
Procedure 
geogebra plaatje


We determine the intersection point of the parabola #y=a_1x^2+b_1x+c_1# and the line #y=a_2x+b_2#. 

Step 1 
First we determine the #x#coordinate of the intersection point by solving the equation \[a_1x^2+b_1x+c_1=a_2x+b_2\] by means of factorization, completing the square or the quadratic formula. 

Step 2 
We determine the #y#coordinate of the intersection point by substituting the obtained #x#coordinate in one of both formulas. Usually it is easier to substitute in the linear formula. 
The last step indicates that given the graphs \[y = x^22\cdot x+3\phantom{xxx}\text{ and }\phantom{xxx} y = x+7\]
intersect each other at two points. Determine the two intersection points.
Give your answer in the form #\left\{\rv{a,b},\rv{c,d}\right\}#, in which #a#, #b#, #c#, #d# are exact numbers.
intersect each other at two points. Determine the two intersection points.
Give your answer in the form #\left\{\rv{a,b},\rv{c,d}\right\}#, in which #a#, #b#, #c#, #d# are exact numbers.
#\left \{\rv{ 1 , 6 } , \rv{ 4 , 11 } \right \} #
The #x#coordinate of a point lying on both parabolas must satisfy
\[x^22\cdot x+3 = x+7\tiny.\]
We solve this equation, after reduction, by factorization.
\[\begin{array}{rcl}
x^23\cdot x4 &=& 0\\
&&\phantom{xxx}\color{blue}{\text{all terms to the left hand side}}\\
\left(x4\right)\cdot \left(x+1\right) &=&0 \\
&&\phantom{xxx}\color{blue}{\text{factorized}}\\
x4 = 0 &\lor& x+1=0 \\
&&\phantom{xxx}\color{blue}{A\cdot B=0 \text{ if and only if }A=0\lor B=0}\\
x=4 &\lor& x=1 \\
&&\phantom{xxx}\color{blue}{\text{constant term to the right hand side}}\\
\end{array}\]
Now we can calculate the corresponding #y#value by entering this #x#value in one of both formulas. In this case it is most convenient to choose the linear function. First we calculate the #y#value at #x=1#.
\[\begin{array}{rcl}
y&= & 1+7 = 6
\end{array}\]
Next we calculate the #y#value at #x=4#.
\[\begin{array}{rcl}
y&=& 4+7 = 11
\end{array}\]
The conclusion is that the #2# points of intersection are given by: \[ \left \{\rv{ 1 , 6 } , \rv{ 4 , 11 } \right \}\tiny. \]
The #x#coordinate of a point lying on both parabolas must satisfy
\[x^22\cdot x+3 = x+7\tiny.\]
We solve this equation, after reduction, by factorization.
\[\begin{array}{rcl}
x^23\cdot x4 &=& 0\\
&&\phantom{xxx}\color{blue}{\text{all terms to the left hand side}}\\
\left(x4\right)\cdot \left(x+1\right) &=&0 \\
&&\phantom{xxx}\color{blue}{\text{factorized}}\\
x4 = 0 &\lor& x+1=0 \\
&&\phantom{xxx}\color{blue}{A\cdot B=0 \text{ if and only if }A=0\lor B=0}\\
x=4 &\lor& x=1 \\
&&\phantom{xxx}\color{blue}{\text{constant term to the right hand side}}\\
\end{array}\]
Now we can calculate the corresponding #y#value by entering this #x#value in one of both formulas. In this case it is most convenient to choose the linear function. First we calculate the #y#value at #x=1#.
\[\begin{array}{rcl}
y&= & 1+7 = 6
\end{array}\]
Next we calculate the #y#value at #x=4#.
\[\begin{array}{rcl}
y&=& 4+7 = 11
\end{array}\]
The conclusion is that the #2# points of intersection are given by: \[ \left \{\rv{ 1 , 6 } , \rv{ 4 , 11 } \right \}\tiny. \]
We see that the calculated intersection points match the intersection points identified in step 1. See the figure below, in which the intersection points are drawn in red.
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