Hishikimi S.
asked • 07/24/16Look for the intersections of the functions
f(x)=2x^2-3x+1
g(x)=-x^2+2x+3
Give your answers as coordinates (a,b)
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2 Answers By Expert Tutors
Nicolas M. answered • 07/25/16
Tutor
5
(1)
Bilingual Tutor Math and Spanish
2x^2+x^2-3x-2x+1-3=0
3x^2 -5x -2 = 0
(3x -2)(x -1) - 0
Zeros: x = 2/3 and x=1 (both x values are the x-intersects values)
Substitute x=2/3 and x=1 in f(x) or g(x) to find theirs associated y values.
Nicolas M.
I made a mistake here:
2x^2+x^2-3x-2x+1-3=0
3x^2 -5x -2 = 0
3x^2 -5x -2 = 0
(3x -2)(x -1) - 0 is not correct
The solution is given by the formulae:
ax^2 + bx + c =0
x = (-b +/- sqrt(b^2 - 4ac))/2a where: a = 3, b=-5 and c=-2
x = (5 +/- sqrt(25 - 4(3)(-2))/6
x = (5 +/- sqrt(25 + 24))/6
x = (5 +/- sqr(49))/6 = (5 +/- 7)/6
There are 2 solutions: X1 = (5+7)/6 and X2 = (5-7)/6
X1 = 2 and X2 = -1/3
Then: 3x^2 -5x -2 = 0 = (x-2)*(3x +1) CORRECT
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07/25/16
Michael J. answered • 07/25/16
Tutor
5
(5)
Mathematical Reasoning and Logic Application
Set f(x)=g(x)
Then solve for x from the equation. When solving for x, move all terms to one side of the equation so that the other side is equal to zero. In the end, you will have a quadratic equation.
After you found your x values, substitute those values into any of the equations to find y.
The way I see it, g(x) looks good to me.
Charles W.
is this correct?
2x^2-3x+1=-x^2+2x+3
2x^2+x^2-3x-2x+1-3=0
3x^2-5x-2=0
(3x^2-6x) (x-2)
3x(x-2) (x-2)=0
What happens next?
Thank you!
Report
07/25/16
Michael J.
You made a mistake in your factoring. Use FOIL method.
3x2 - 5x - 2 = 0
(3x + 1)(x - 2) = 0 ----> correct factoring
Report
07/25/16
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Nicolas M.
07/24/16