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In this lesson, we will learn how to find the intersection points of two functions, using a graphical or algebraic approach to solve systems of equations where one or both are nonlinear.

Q1:

Find the set of points of intersection of the graphs of π¦ = 3 π₯ and π₯ + π¦ = 4 0 2 2 .

Q2:

Find the set of points of intersection of the graphs of π₯ = π¦ and π₯ + π¦ = 3 2 2 2 .

Q3:

Find the set of points of intersection of the graphs of π₯ = π¦ and π₯ + π¦ = 1 6 2 2 2 .

Q4:

Find the set of points of intersection of the graphs of π₯ = π¦ and π₯ + π¦ = 2 0 0 2 2 .

Q5:

The given figure shows the graphs of the functions π ( π₯ ) = 2 π₯ β 2 and π ( π₯ ) = ( π₯ ) l n . What are the points where π ( π₯ ) = π ( π₯ ) ?

Q6:

The given figure shows the graphs of the functions π ( π₯ ) = 6 and π ( π₯ ) = 3 π₯ β 3 π₯ 2 . What are the points where π ( π₯ ) = π ( π₯ ) ?

Q7:

Answer the following questions.

Determine the equation of the line that passes through the points (2, 0) and (1, 2).

Determine the equation of the line that passes through the points ( β 1 , 0 ) and ( 0 , β 2 ) .

Hence, do the two lines intersect? If yes, state the point of intersection.

Q8:

The graphs of π 1 and π 2 intersect at the point ( π , 2 ) , where π ( π₯ ) = 2 1 π₯ and π ( π₯ ) = 3 β π₯ 2 . Find the set of possible values of π .

Q9:

Answer the following questions for the functions π¦ = π₯ β 3 π₯ β 4 ο¨ and π¦ = π₯ + 1 .

Complete the table of values for π¦ = π₯ β 3 π₯ β 4 ο¨ .

Complete the table of values for π¦ = π₯ + 1 .

Use the tables of values to determine an intersection point of the two graphs.

By extending the table up to π₯ = 8 , check if there are other intersection points. If so, find their coordinates.

Q10:

The given figure shows the graphs of the functions π ( π₯ ) = 3 π₯ β 3 and π ( π₯ ) = π₯ + 4 π₯ β 5 2 . What are the points where π ( π₯ ) = π ( π₯ ) ?

Q11:

Find all values of π₯ where π ( π₯ ) = π‘ ( π₯ ) , given π ( π₯ ) = ( π₯ + 3 4 ) 2 and π‘ ( π₯ ) = π₯ + 3 4 .

Q12:

The given figure shows the graphs of the functions π ( π₯ ) = 5 π₯ β 4 and π ( π₯ ) = β π₯ + 8 . What is the point where π ( π₯ ) = π ( π₯ ) ?

Q13:

Determine the equation of the line that passes through the points ( β 2 , β 1 ) and ( 0 , 3 ) .

Determine the equation of the line that passes through the points ( β 2 , 4 ) and ( β 1 , 1 ) .

Q14:

Determine the equation of the line that passes through the points ( 3 , 1 ) and ( 5 , 3 ) .

Determine the equation of the line that passes through the points ( 2 , 4 ) and ( 3 , 3 ) .

Q15:

The given diagram shows the graph of π ( π₯ ) = οΉ π₯ β 7 5 π₯ ο 5 0 3 together with three lines of the form π¦ = π β π₯ .

By solving π ( π₯ ) = β π₯ , determine the coordinates of points π΄ and π΅ .

Note that from π ( 2 ) = β 7 1 2 5 , we know that line π¦ = 2 β 7 1 2 5 β π₯ meets π¦ = π ( π₯ ) in the point οΌ 2 , β 7 1 2 5 ο , so ( π₯ β 2 ) is a factor of the cubic polynomial π ( π₯ ) β οΌ 2 β 7 1 2 5 β π₯ ο . Find the π₯ -coordinates of the other two intersections.

Simplify the expression that states that the average rate of change of π from π₯ = π to π₯ = π is β 1 to a quadratic equation in π that has coefficients involving π .

The π₯ -coordinate π of point πΆ is one where there is just one π for which the average rate of change of π is β 1 . By finding the discriminant of the quadratic expression find π and π .

For what values of π are there no points π so that the average rate of change of π from π₯ = π to π₯ = π is β 1 ?

Q16:

Find all the possible values of π₯ satisfying π ( π₯ ) = π₯ + 9 π₯ + 2 2 and π ( π₯ ) = β 1 6 given π₯ β β€ .

Q17:

The given figure shows the graphs of the functions π ( π₯ ) = 4 π₯ β 2 and π ( π₯ ) = β 2 π₯ + 4 . What is the point where π ( π₯ ) = π ( π₯ ) ?

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