A polynomial and a linear function intersect only once if the linear function is a tangent line.
Find b such that f(x) and g(x) intersect at one point only.
f(x) = 3x^2 + 5x + 7
g(x) = bx + 3
I set them equal
3x^2 + 5x + 7 = bx + 3
3x^2 + (5-b)x + 4 = 0
-(5-b) + - sqrt b^2 - 10b + 25 - (48)
-5 + b + - sqrt b2 -10b -23 / 6 I don't know what to do next the answer is suppose to be 5 + - 4 sqrt 3
How did they get rid of the "b" in the sqrt
I would love some help
You have to find a tangent line to f(x)=3x^2 + 5x + 7 such that it has the form bx+3.
The slope of the tangent line is given by f'(x)=6x+5, so that at a point y the slope is 6y+5 and the tangent line is given by h(x)=(6y+5)x+c.
Also at that point y the value of the tangent is equal to the value of the polynomial in that point so that you have h(y)=f(y). Now you look for the values of y such that c=3, then you have h(y)=(6y+5)y+3 and f(y)=3y^2+5y+7. These should be equal, so you find y=+/- sqrt(4/3).
So b=6(+/- sqrt(4/3))+5 = 5+/-sqrt(48) = 5+/-4sqrt(3).