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
Inactive Tutor answered • 07/25/16
Tutor
New to Wyzant
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.
Inactive Tutor
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
Inactive Tutor answered • 07/25/16
Tutor
New to Wyzant
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
Inactive Tutor
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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Inactive Tutor
07/24/16