Find the point(s) of intersection of the parabola with equation y = x^2 - 5x + 4 and the line with equation y = 2x - 2
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At the point of intersection of the parabola y = x^2 - 5x + 4 and the line y = 2x - 2, the value of x and y is equal.
So we can equate the two and solve for x.
x^2 - 5x + 4 = 2x - 2
=> x^2 - 7x + 6 = 0
=> x^2 - 6x - x + 6 = 0
=> x(x-6) - 1(x - 6) = 0
=> (x - 1)(x -6) = 0
So we have x = 1 and 6
The value of y for x = 1 is 0 and for x = 6, y = 10.
Therefore the point of intersection are ( 1, 0) and (6, 10)
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calendarEducator since 2008
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starTop subjects are Math, Science, and Social Sciences
Given the parabola y= x^2 - 5x + 4 and the line y= 2x-2
We need to find the intersection points of the parabola and the line.
The intersection points are the values of x and y such that:
the parabola y = the line y
==> x^2 - 5x + 4 = 2x -2
We will combine like terms and solve for x.
==> x^2 - 7x + 6 = 0
Now we will factor.
==> (x-6)(x-1) = 0
==> x1= 6 ==> y1= 2*6-2 = 10
==> x2 = 1==> y2= 2*1 -2 = 0
Then we have two intersection points:
The intersection points of the parabola and the line are (6, 10) and (1,0)
Q: Find the point(s) of intersection of the parabola with equation y = x2 - 5x + 4 and the line with equation y = 2x - 2.
A:
y = x^2-5x+4 ...(1).
y = =2x-2.......(2).
The y coordinates at the point of intersection is same.
So we equate the right sides of (1) and (2):
x^2-5x+4 = 2x-2. We subtract x-2 from both sides and get:
=> x^2-5x+4-2x+2 = 0.
=> x^2-7x+6 = 0.
=> (x-6)(x-1) = 0.
So x-2 = 0, or x-1 = 0.
=> x= 6, or x= 1.
Therefore we put x= 6, or x= 1 in equation (2), y = 2x-2. to obtain ordinates :
When x = 6, y = 2*6-2 = 12-2 = 10.
When x= 1, y = 2*1-2 = 0.
Therefore (6,10) and (1,0) are two points of intersection of the parabola y = x^2-5x+4 and the line y = 2x-2 .
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