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Find the distance between the points of intersection of the graphs of the functions.

1) y=x^2-3x+4 and y=x+1
2) y=x^2-4 and y=2x-4

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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