Intersection points of parabolas

Quadratic equations: Intersection points of parabolas

Intersection points of parabolas

Two parabolas can also intersect at zero, one or two points.

We will now take a look at how to find these intersection points.

Intersection points of two parabolas

	Procedure	geogebra plaatje
	We determine the intersection point of the parabolas #y=a_1x^2+b_1x+c_1# and #y=a_2x^2+b_2x +c_2#.
Step 1	We first determine the #x#-coordinate of the intersection point by solving the equation \[a_1x^2+b_1x+c_1=a_2x^2+b_2x+c_2\] by means of reduction and if needed 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.

Number of intersection points and graphJust as with the intersection point of a parabola and a line, we can see from the graph how many intersection points the two parabolas have. Another option to find the number of intersection points is to solve the equation in step 1. The number of intersection points can be either zero, one or two.

Linear equationWhen solving the equation from step 1, you can end up with a linear equation if the values of #a_1# and #a_2# are equal. In that case, the parabolas have either zero or one intersection points.

Example

Parabolas #y=2x^2+5x+3# and #y=2x^2+3x-1#

Step 1

To find the #x#-coordinate of the intersection point we solve the equation #2x^2+5x+3=2x^2+3x-1#. This is done as follows.

\[\begin{array}{rcl}2x^2+5x+3&=&2x^2+3x-1 \\ &&\phantom{xx}\blue{\text{the equation we need to solve}}\\5x+3&=&3x-1 \\ &&\phantom{xx}\blue{\text{both sides minus }2x^2}\\ 2x+3&=& -1 \\ &&\phantom{xx}\blue{\text{both sides minus }3x}\\ 2x&=& -4 \\ &&\phantom{xx}\blue{\text{both sides minus }3}\\ x&=& -2 \\&&\phantom{xx}\blue{\text{both sides divided by }2} \end{array}\]

Hence, the #x#-coordinate of the intersection point is #x=-2#.

Step 2

To find the #y#-coordinate of the intersection point we substitute #x=-2# in #y=2x^2+5x+3#. This gives us: \[y=2 \cdot \left(-2\right)^2+5\cdot -2 +3=1\]

Hence, the #y#-coordinate of the intersection point is #y=1#.

The intersection point of #y=2x^2+5x+3# and #y=2x^2+3x-1# is #\rv{-2,1}#.

Calculate the intersection points of the parabolas given by
\[y = x^2-4\cdot x+4\phantom{xxx}\text{ and }\phantom{xxx} y = x^2+4\cdot x+1\tiny.\]
Give your answer in the form of

#none# #\phantom{xxwwxx}# if there is no intersection,
#\left\{\rv{a,b}\right\}\phantom{xxxww}# if there is one intersection, and
#\left\{\rv{a,b},\rv{c,d}\right\}\phantom{x}# if there are two intersections,

in which #a#, #b#, #c#, and #d# are exact numbers.

#\left\{\rv{{{3}\over{8}},{{169}\over{64}}}\right\}#

Step 1

The #x#-coordinate of a point on both parabolas must satisfy
\[x^2-4\cdot x+4 = x^2+4\cdot x+1\tiny.\]

We solve this equation, after reducing it to #0#.
\[\begin{array}{rcl}
x^2-4\cdot x+4&=&x^2+4\cdot x+1 \\
&&\phantom{xxx}\blue{\text{original equation}}\\
3-8\cdot x &=& 0\\
&&\phantom{xxx}\blue{\text{all terms moved to the left hand side}}\\
\displaystyle x&=&\displaystyle {{3}\over{8}} \\
&&\phantom{xxx}\blue{\text{equation solved}}
\end{array}\]

Step 2

We calculate the corresponding #y#-value by substituting #x={{3}\over{8}}# in one of the formulas for the parabolas. Here, we will substitute the #x#-value in the first formula.
\[\begin{array}{rcl}
y&=& \left({{3}\over{8}}\right)^2 -4 \cdot \left({{3}\over{8}}\right) + 4\\
&=&\displaystyle {{169}\over{64}}
\end{array}\]

The conclusion is that there is #1# intersection point, given by : \[ \left\{\rv{{{3}\over{8}},{{169}\over{64}}}\right\}\tiny. \]

In the figure below the parabola #y=x^2-4\cdot x+4# is drawn in blue and #y=x^2+4\cdot x+1# in green. The intersection point is drawn in red.

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