f(x) = 3x^2 + 5x + 7. g(x) = bx + 3. At what b will these functions intersect?

Question

Is there a way to solve this algebraically without having to graph?

Answer 1

3x^2 + 5x + 7 = bx + 3
3x^2 + (5x-bx) + 4 = 0
3x^2 + (5-b)x + 4 = 0
Try b = 12.
The points of intersection would be
(1,15) and (4/3,19)
b = -2 is another solution.
The points of intersection (for b = -2) are
(-1,5) and (-4/3,35/3)

Answer 2

yes
1) Equal both equations to each other
3x^2 + 5x +7 = bx +3
2)get the bx alone in one side
bx= 3x^2 + 5x +7 -3
3)divide by x
b = 3x + 5 + 4/x
If you graphed it..it doesnt have a point of intersection...f(x) is a parabola and g(x) is a line with zero slope

Answer 3

5x^4 - 14x ^3 + 18x^2 + 40x + 16 = (x^2 -4x +8)(ax^2 +bx+c) boost RHS and simplify ax^4 - 4ax^3 + 8ax^2 + bx^3 - 4bx^2 + 8bx + cx^2 - 4cx + 8c =>x^4(a) -x^3(4a - b) + x^2(8a-4b + c) + x(8b -4c) + 8c comparing with coreesponding words in LHS a = 5 -------------------eqn(a million) 4a - b = 14------------eqn(2) 8a-4b+c = 18------------eqn(3) 8b-4c=40--------------eqn(4) 8c = 16-----------------eqn(5) from(a million) a = 5: from (5) c = 2 plugging in a cost in (2) --4*5 - b = 14 <=>20 - b = 14 <-> b = 6 so a = 5: b = 6 : c = 2 so 5x^4 - 14x ^3 + 18x^2 + 40x + 16 = (x^2 -4x +8)(5x^2 +6x+2) to locate all recommendations equate RHS = 0 (x^2 -4x +8)(5x^2 +6x+2) = 0 even as x^2 - 4x + 8 = 0, making use of quadratic formulation x = [4 ± sqrt(16 -32)]/2 = [4± sqrt(-16)]/2 = [4 ± 4i]/2 x = 2 ± 2i x = 2+ 2i or 2 -2i even as 5x^2 + 6x + 2 = 0 x = [-6±sqrt(36 -40)]/10 x = [-6 ± sqrt(-4)]/10 = [-6 ± 2i]/10 x = (a million/5)(-3 + i) or (a million/5)(-3 - i)

Answer 4

I assume that this is a question about finding the slope of a line tangent to a curve. What you need to do is take the derivative of f(x) then set that as b in g(x). Then solve for x when g(x)=0 and that will give you your x inercept. After that all you need to do is divide your y intercept by your x intercept and that should give you your answer. I think.

Answer 5

This is either a trick question, or you have missed important information. For g(x), if you are allowed to choose ANY b value at all...that is the slope, so you can choose many, many b values that will intersect f(x).

You could make it intersect only once, or twice if you wanted.

Answer 6

At b = 12. f(x) is a quadratic functiona and g(x) is linear. The only place they can ever meet is when x = 1 or 4/3.

Answer 7

there is a way to do this algebraically, but it involves derivatives.

Answer 8

yea there is but i just cant rember becuz i dont go to school anymore ummgo to www.enise,com

Answer 9

yeah im sorry I missed this day in class, but good luck !

Answer 10

there is put it would take forever. graphing it would be much easier