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Given the gradient and finding the equation? - HSN forum

# Given the gradient and finding the equation?

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### #1louisa567

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Posted 15 November 2006 - 08:51 PM

Im completely stuck on a few homework questions we have been given. I dont know where I am going wrong:

Given its gradient, find the equation of the tangent to this parabola.

y= x squared, m = 2 ?

Im completely lost but when I do b squared - 4ac i get a = 1, b= -2 and c = -1. Could this be where I am going wrong?

The answer in the book is y = 2x - 1. How do they get this?

### #2John

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Posted 15 November 2006 - 08:56 PM

To find the equation of the tangent to a curve you narmally differentiate.

So do that, you know that dy/dx = 2 so find what dy/dx is and solve it for 2.

So after that you have the x co-ordinate, plug that into the equation of the curve and use the answer from that for the y co-ordinate.

After that use the formula y-b=m(x-a)

Does that make any sense to you?

### #3louisa567

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Posted 15 November 2006 - 09:06 PM

Yea that make sense John thanks but its just we are doing the quadratics topic so our teacher wants us to use b^2 - 4ac etc.

### #4John

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Posted 15 November 2006 - 09:12 PM

the discrminant (b2-4ac) only tells you the nature of the roots of the quadratic

### #5The Wedge Effect

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Posted 15 November 2006 - 09:55 PM

It also tells you when the line touches the parabola, when it doesn't pass through the parabola, and finally, when it passes through the parabola at two points, with bē-4ac=0, bē-4ac<0 and bē-4ac>0, respectively if you make the equation of the parabola equal the line equation, and equate to zero.

### #6louisa567

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Posted 16 November 2006 - 08:18 PM

Thanks wedge effect! I understand it now.

### #7The Wedge Effect

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Posted 16 November 2006 - 08:26 PM

That's the Wedge Effect for you.

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