FIND POINTS OF INTERSECTION OF PARABOLA AND LINE

Without using technology, find the coordinates of the point(s) of intersection of:

Problem 1 :

y = x2 - 2x + 8 and y = x + 6

Solution:

y = x2 - 2x + 8 ---> (1)

y = x + 6 ---> (2)

x2 - 2x + 8 = x + 6

x2 - 2x - x + 8 - 6 = 0

x2 - 3x + 2 = 0

(x - 1) (x - 2) = 0

x = 1 or x = 2

Substitute x = 1 in equation (1), we get

y = 12 - 2(1) + 8

y = 7

Substitute x = 2 in equation (2), we get

y = 2 + 6

y = 8

Therefore, intersection points are (1, 7) and (2, 8).

Problem 2 :

y = -x2 + 3x + 9 and y = 2x - 3

Solution:

y = -x2 + 3x + 9 ---> (1)

y = 2x - 3 ---> (2)

-x2 + 3x + 9 = 2x - 3

x2 + 2x - 3x - 3 - 9 = 0

x2 - x - 12 = 0

(x - 4) (x + 3) = 0

x = 4 or x = -3

Substitute x = 4 in equation (1), we get

y = -(4)2 + 3(4) + 9

y = -16 + 12 + 9

y = 5

Substitute x = -3 in equation (2), we get

y = 2(-3) - 3

y = -6 - 3

y = -9

Therefore, intersection points are (4, 5) and (-3, -9).

Problem 3 :

y = x2 - 4x + 3 and y = 2x - 6

Solution:

y = x2 - 4x + 3 ---> (1)

y = 2x - 6 ---> (2)

x2 - 4x + 3 = 2x - 6

x2 - 4x - 2x + 3 + 6 = 0

x2 - 6x + 9 = 0

(x - 3) (x - 3) = 0

x = 3

Substitute x = 3 in equation (1), we get

y = (3)2 - 4(3) + 3

y = 9 - 12 + 3

y = 0

Therefore, intersection point is (3, 0).

Problem 4 :

y = -x2 + 4x - 7 and y = 5x - 4

Solution:

y = -x2 + 4x - 7 ---> (1)

y = 5x - 4 ---> (2)

-x2 + 4x - 7 = 5x - 4

x2 + 5x - 4x - 4 + 7 = 0

x2 + x + 3 = 0

This expression has no real factors. So, the curves do not intersect.

Problem 5 :

For which value of c is the line y = 3x + c a tangent to the parabola with equation y = x2 - 5x + 7 ?

Solution:

y = x2 - 5x + 7

Slope (dy/dx) = 2x - 5

y = 3x + c

where 3 is the slope and c is the y-intercept

2x - 5 = 3

2x = 8

x = 4

Substitute x = 4 in above equation

y =  (4)2 - 5(4) + 7

y = 16 - 20 + 7

y = 3

Therefore, the point of intersection is at (4, 3).

y = 3x + c

3 = 3(4) + c

3 = 12 + c

c = -9

Problem 6 :

Find the values of m for which the lines y = mx - 2 are tangents to the curve with equation y = x2 - 4x + 2.

Solution:

y = x2 - 4x + 2 ---> (1)

m = Slope (dy/dx) = 2x - 4

y = mx - 2

y = (2x - 4)x - 2

y = 2x2 - 4x - 2 ---> (2)

(1) = (2)

x2 - 4x + 2 = 2x2 - 4x - 2

x2 - 4 = 0

x2 = 4

x = 2 or -2

If x = 2,

m = 2x - 4

m = 2(2) - 4

m = 4 - 4

m = 0

If x = -2

m = 2x - 4

m = 2(-2) - 4

m = -4 - 4

m = -8

Hence, m values are 0 and -8.

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