Question 914436: Given these two equations : f(x)=x^2-6x+9 and g(x)=-3x-5, change the slope of the line so that it will intersect the parabola in two locations.
What I have done so far:
Let f(x)=g(x)
x^2-6x+9=-3x-5
x^2-3x+14=0
Then I calculated the discriminant to see how many solutions it would have.
D=b^2-4ac
D=(-3)^2-4(1)(14)
D=9-56
D=-47
Since the discriminant is -47, therefore there are no real solutions to x^2-3x+14=0, which means that f(x) and g(x) never intersect.
What I did for the second part was represent the slope of the second equation with the variable "m".
f(x)=x^2-6x+9
g(x)=mx-5
Let f(x)=g(x)
x^2-6x+9=mx-5
x^2-6x-mx+9+5=0
x^2-(6-m)x+14=0
a=1, b=-6-m, c=14
Sub a, b and c into discriminant equation
D=b^2-4ac
D=(-6-m)(-6-m)-4(1)(14)
D=36+6m+6m+m^2-56
D=m^2+12m-20
This cannot be factored, so I don't know what to do next!
Thanks
Answer by ewatrrr(24785)   (Show Source):
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