- #1
Painguy
- 120
- 0
1. show that there is no line through the point (2,7) that is tangent to the parabola y =x^2 +x
2. y-y1=m(x-x1)
3.
[itex]y'=f'(x)=2x+1[/itex]
[itex]m1=2x +1[/itex]
[itex]m1=2(2) +1 =5[/itex]
[itex]m2=((x^2 +x)-7)/(x-2)[/itex]
[itex]m1=m2?[/itex]
[itex]((x^2 +x)-7)/(x-2)=5[/itex]
[itex]x^2-4x+3=0[/itex]
[itex]2^2-4(2)+3=/=0[/itex]
[itex]-1=0[/itex]
I'm thinking that i would compare the slope of the line passing through (2,7) to the general slope of the parabola. The other thing i believe i could do is just plug 2 into y=x^2 +x. that would give me 6 which does not equal 7, but that does not involve using any of the material we're covering in class so that is out of the question.
2. y-y1=m(x-x1)
3.
[itex]y'=f'(x)=2x+1[/itex]
[itex]m1=2x +1[/itex]
[itex]m1=2(2) +1 =5[/itex]
[itex]m2=((x^2 +x)-7)/(x-2)[/itex]
[itex]m1=m2?[/itex]
[itex]((x^2 +x)-7)/(x-2)=5[/itex]
[itex]x^2-4x+3=0[/itex]
[itex]2^2-4(2)+3=/=0[/itex]
[itex]-1=0[/itex]
I'm thinking that i would compare the slope of the line passing through (2,7) to the general slope of the parabola. The other thing i believe i could do is just plug 2 into y=x^2 +x. that would give me 6 which does not equal 7, but that does not involve using any of the material we're covering in class so that is out of the question.