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Ksju [112]
2 years ago
5

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

Mathematics
2 answers:
svlad2 [7]2 years ago
8 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.

Viefleur [7K]2 years ago
7 0

Answer:

Find the value of x and y in coordinate form, that'll be the point of intersection.

Question 1

{ \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 }}

Therefore, points of intersection are two

Answer: <u> </u><u>(</u><u>3</u><u>,</u><u> </u><u>4</u><u>)</u><u> </u><u>and</u><u> </u><u>(</u><u>1</u><u>,</u><u> </u><u>2</u><u>)</u>

Question 2:

Following the steps as in question 1

{ \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}}}

Answer: <u>(2, 0)</u>

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