Question

change the slope of the line h(x)=-3x-5 so that it does not intersect the parabola q(X)= 3x^2+4x-2

"slope" h(x) = -3x-5
"parabola" q(x)= 3x^2 +4x-2

Answer 1

@Loser66 help plz

Answer 2

@dpasingh

Answer 3

@dpasingh

Answer 4

@ganeshie8 @Mashy

Answer 5

@BTaylor

Answer 6

Solve by using h(x) = q(x) in order to find the intersection of the parabola by line.
The value of x would be the pint of intersection.

Answer 7

I have found that these two already intersect due to the fact that when you graph them you can identify that, but whats troubling is how you can change the slope to not make them intercept..

Answer 8

agree with dpasingh, when let them equal, you have a quadratic. Let do it
3x^2 + 4x -2 = ax + 5, and we have to find a such that 2 lines intersect, right?

Answer 9

3x^2+4x -ax-7 =0
3x^2 +(4-a)x -7 =0

Answer 10

to get the intersect point, we need \(\triangle = 0\), it means \(\sqrt{(4-a)^2 -4 (3)*(-7)}=0\)
or (4-a)^2 +84 =0
solve for a, that is the condition of slope

Answer 11

how about if you don't want them to intersect?

Answer 12

If they intersect at 2 points, that mean \(triangle >0)
if they don't intersect, then \(\triangle <0) ,
those are the condition for the quadratic has the solutions ( real solution, no solution, double root solution)

Answer 13

Not sure whether you got me or not.
let me interpret the stuff
for a quadratic ax^2 +bx +c =0
to have 2 solutions, we need \(\triangle >0\)
to have "no real solutions, we need \(\triangle <0\)
to have one solution, we need \(\triangle =0\)
to this particular problem, you get the quadratic by let g(x) = h(x) in which slope is an unknown a.

Answer 14

and from it , you have the condition of a, such that \(\triangle >0, <0; =0\) to have 2 curves intersect at 2points, 0 point, and 1 point respectively

Answer 15

aaaahhhh, is there any one understand what I am trying to say?

Answer 16

I f you do not want to isee them intersect, then take change the value of x which is other than the point of intersection of x

Answer 17

I understand that part but what I don't understand is the part where (3x^2+4x-2=ax+5) .. the equation that the question have is -3x-5 so how did it change to ax+5 ..?

Answer 18

because you don't know what is the value of the slope (the number stay in the front of x) make 2 curves intersect, and you have to solve for it, when a =-3, do they intersect? you don't know, you have to test
what if it is 4?5? 6? and you have to test one by one until you find out which values make them intersect. right?

Answer 19

yes

Answer 20

so, I just put it in general to solve for a so that I know with what the value of a, 2 curves intersect