Equation of both lines that are tangent to the graph y=x^2

  • Thread starter francis21
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  • #1
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Homework Statement


Question: Determine the equations of both lines that are tangent to the graph of f(x) = x2 and pass through point (1,-3).

Homework Equations


Some of the equations that I could use for this problem are:
y-y1=m(x-x1) (Point-slope Equation)

the derivative of the function f(x) from first principles (the limit of the difference quotient, as h approaches 0)

The Attempt at a Solution



First, I took the derivative of the function x2.

As a result,

f`(x) = 2x

But I'm not sure on how to go from here.
 

Answers and Replies

  • #2
danago
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Lets say that the line is tangent to the curve at the point with coordinates (p,p2). Using the definition of the gradient of a line along with the fact that we know it passes through (1,-3) as well as (p,p2):

[tex]m = \frac{\Delta y}{\Delta x} = \frac{p^2+3}{p-1}[/tex]

Since we know that the line is tangent to the curve with equation y=x2, whose derivative, as you noted, is y'=2x, it should be clear that the slope of the line can also be expressed as m=2p, since we originally defined the point (p,p2) as the point of tangency.

Can you go from there?
 
  • #3
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Hey, thanks for the reply.

But is there another approach to this problem, without the use of gradients (since it has not been taught yet in my calculus class)? Sorry about that. That's why, it's hard to recognize what you're trying to explain.
 
  • #4
vela
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By gradient, danago just means slope.
 

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