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Sagot :
Answer:
Find the value of x and y in coordinate form, that'll be the point of intersection.
Question 1
[tex]{ \rm{y = {x}^{2} - 3x + 4 }} \\ { \boxed{ \tt{but \: y = x + 1 \: }}} \\ \\ { \rm{(x + 1) = {x}^{2} - 3x + 4 }} \\ \\ { \rm{ {x}^{2} - 4x + 3 = 0 }} \\ \\ { \rm{(x - 3)(x - 1) = 0}} \\ \\ { \boxed{ \rm{x_{1} = 3 \: \: and \: \: x _{2} = 1}}} \\ \\ { \boxed{ \tt{remember \: y = x + 1}}} \\ \\ { \rm{y _{1} = 4 \: \: and \: \: y _{2} = 2 }}[/tex]
Therefore, points of intersection are two
Answer: (3, 4) and (1, 2)
Question 2:
Following the steps as in question 1
[tex]{ \rm{y = {x}^{2} - 4}} \\ \\{ \rm{2x - 4 = {x}^{2} - 4}} \\ \\ { \rm{ {x}^{2} = 2x }} \\ \\ { \boxed{ \rm{x = 2}}} \\ { \tt{remember : \: y = 2x - 4 }} \\ { \boxed{ \rm{y = 0}}}[/tex]
Answer: (2, 0)
Answer:
Below in bold.
Step-by-step explanation:
1) y=x^2-3x+4 and y=x+1
Using substitution for y :
x + 1 = x^2 - 3x + 4
x^2 - 4x + 3 = 0
(x - 3)(x - 1) = 0
x = 1, 3.
If x = 1, y = 1-3+4 = 2 and
if x = 3, y = 9-9+4 = 4.
So the points of intersection are (1, 2) and (3, 4)
Distance between them = √[(3-1)^2 + (4-2)^2 ] = √8.
2) y=x^2-4 and y=2x-4
2x - 4 = x^2 - 4
x^2 - 2x = 0
x(x - 2) = 0
x = 0, 2
When x = 0, y = -4 and
when x = 2, y = 0
So the points are (0,-4) and (2, 0)
So distance between the 2 points = √[(2-0)^2 + (0--4)^2)] = √20.
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