Without using technology, find the coordinates of the point(s) of intersection of:
Problem 1 :
y = x^{2} - 2x + 8 and y = x + 6
Solution:
y = x^{2} - 2x + 8 ---> (1)
y = x + 6 ---> (2)
x^{2} - 2x + 8 = x + 6
x^{2} - 2x - x + 8 - 6 = 0
x^{2} - 3x + 2 = 0
(x - 1) (x - 2) = 0
x = 1 or x = 2
Substitute x = 1 in equation (1), we get
y = 1^{2} - 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 = -x^{2} + 3x + 9 and y = 2x - 3
Solution:
y = -x^{2} + 3x + 9 ---> (1)
y = 2x - 3 ---> (2)
-x^{2} + 3x + 9 = 2x - 3
x^{2} + 2x - 3x - 3 - 9 = 0
x^{2} - 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 = x^{2} - 4x + 3 and y = 2x - 6
Solution:
y = x^{2} - 4x + 3 ---> (1)
y = 2x - 6 ---> (2)
x^{2} - 4x + 3 = 2x - 6
x^{2} - 4x - 2x + 3 + 6 = 0
x^{2} - 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 = -x^{2} + 4x - 7 and y = 5x - 4
Solution:
y = -x^{2} + 4x - 7 ---> (1)
y = 5x - 4 ---> (2)
-x^{2} + 4x - 7 = 5x - 4
x^{2} + 5x - 4x - 4 + 7 = 0
x^{2} + 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 = x^{2} - 5x + 7 ?
Solution:
y = x^{2} - 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 = x^{2} - 4x + 2.
Solution:
y = x^{2} - 4x + 2 ---> (1)
m = Slope (dy/dx) = 2x - 4
y = mx - 2
y = (2x - 4)x - 2
y = 2x^{2} - 4x - 2 ---> (2)
(1) = (2)
x^{2} - 4x + 2 = 2x^{2} - 4x - 2
x^{2} - 4 = 0
x^{2} = 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.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM