Finding Tangent Lines of f(x) Algebraically

[Quadratic]

Ok, I've found the answer to these questions, but I did so in more of a trial and error way. The question is:

f(x) = 2x^2 + x

Find the two tangent lines to the curve, which both pass through the point (2,-3).

So, I tried using y=mx+b = 2x^2 + x, where m = f'(x) = 2x + 1, thus:
(2x+1)x + b = 2x^2 + x. I worked this out to find an extraneous answer, and one correct tangent of y = -x - 1. I know the other answer has a slope of 11, through trial and error, but there's got to be an easier way.

Any thoughts on how to do this algebraically?

 

[benorin]

Ok, so f(x) = 2x^2 + x\Rightarrow f^{\prime} (x) = 4x + 1

We wish to find the two tangent lines to the curve, which both pass through the point (2,-3). Use the point-slope formula instead, that is, use

y-y_0=m(x-x_0)

Since the tangent line(s) pass through the point (2,-3), we may take (x_0,y_0)=(2,-3), hence

y+3=m(x-2)

but also, if (a,f(a)) is the point at which the line is tangent to the curve, we require that, m=f^{\prime} (a) = 4a + 1 and that (a,f(a)) be a point on the tangent line itself, hence we require that

f(a)+3=f^{\prime} (a)(a-2) \Rightarrow 2a^2+a+3=(4a+1)(a-2)

hold. The two solutions for a give the two tangent lines.

 

[Quadratic]

Ahh, I see. It looks like I wrote the function down wrong (should be x^2 + x), but thanks for the insight.

 

[benorin]

That you may check your work: the values of a that work for f(x)=x^2+x are a=2\pm 3 = -1,5 and the other tangent line is y=11x-25