Intersection point between a line with slope m and f(x)

[cir]

Homework Statement

f(x) = 2x²-5x-12 is given;
part a: find derivative of f(x) using first principles,
part b: find the rate of change of f(x) at x=1,
part c: the points at which the line through (1, -15) with slope m cuts the graph of f(x),
part d: the value of m such that the points of intersection found in (c) are coincident,
part e: the equation of the tangent to the graph of f(x) at the point (1, -15)

2. The attempt at a solution
I found the derivative first, using first principles, it is f'(x) = 4x - 5

Then for part b, the rate of change of f(x) at x = 1, it's the f'(x) value at x = 1, which is -1, the slope of tangent.

I am stuck at part c. I first got the line equation; y-y1=m(x-x1) using point (1, -15), which gives y = mx-m-15

So to find the intersection points with this line and f(x), I tried to equate them both and solve,

mx-m-15 = 2x²-5x-12, and end up at the following,

2x²-(5-m)x+(m+3)=0 how do I solve this? or is my approach wrong?

And what is that part d is asking? I didn't understand it.

Thanks!

 

[rock.freak667]

You are correct thus far.

Part d is saying that the equation you found 2x²-(5-m)x+(m+3)=0 has only one root.

 

[rl.bhat]

Hi cir, welcome to PF.
2x2-(5-m)x+(m+3)=0
This equation should be
2x2-(5+m)x+(m+3)=0
If ax^2 + bx + c = 0, then x = [-b (+ or -)(b^2 - 4ac)^1/2 ]/2a
Use this formula and solve for x.

 

[cir]

@rl.bhat

Thanks. Just saw the sign error. Now able to solve using factorization too.

mx-m-15=2x²-5x-12

2x²-5x-mx+m+3=0

2x²-5x+3-m(x-1)=0

2x(x-1)-3(x-1)-m(x-1)=0

(2x-3-m)(x-1)=0

x = 1 ; x = (m+3)/2

y= -15 ; y = ((m²+m)/2)-15

intersecting points therefore: pt (1, -15), pt ((m+3)/2, ((m²+m)/2)-15) ;answer correct according to textbook answers.


@rock.freak667

I still don't get it , now that I have found the intersection points, how do I find the value of m such that the points of intersection found in (c) are coincident? how do I know it's just one root (cause I did get two intersection points, (1, -15) and the other in terms of m)?

My guess is since the pt (1, -15) is on both lines, thus coincident to both lines, I could equate the x values of the other intersection point found in terms of slope m ((m+3)/2, (m²+m)/2-15)

Thus equating x coordinates gives (m+3)/2 = 1 and solving for m, m = -1 (equating y coordinates gives m = 0)

m = -1 is the correct answer in book, but I still don't understand this part completely. some further explanation would be great

 

[rl.bhat]

The tangent to the curve touches the curve at one point.
You have already found f'(x) = m = (4x - 5). Put x = 1. You get m = -1.

 

[cir]

got it, thnks.