(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

f(x) = 2x^{2}-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 = 2x^{2}-5x-12, and end up at the following,

2x^{2}-(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!

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# Homework Help: Intersection point between a line with slope m and f(x)

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