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- anonymous

A curve has a gradient function px^2-4x where p is a constant. The tangent to the curve at the point (1,3) is parallel to the straight line y+x-5=0 ?
Can anyone explain gradient function and what is tangent to the point.?? Please explain the concept. I dont understand it.

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- anonymous

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- sasogeek

a gradient tells the slope of a line. the gradient function of a curve allows you to know the slope of a tangent at any given point.
a tangent is a line that just touches a curve at a point :)

- sasogeek

|dw:1337323457233:dw|

- sasogeek

umm the gradient function \(\ f'(x)=px^2-4x \) according to the question

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- sasogeek

you might want to check your solution again :/ seeing as you placed f(x) to be the gradient function and f'(x) to be the second derivative...

- anonymous

So If it is gradient function?It already refer to the graph?? Still didnt get it?

- sasogeek

unless i'm wrong and mistaking, you might want to correct me then :/ idk

- anonymous

@sasogeek U r right. because the answer is p=3

- sasogeek

how did you solve it maths911? just being curious

- anonymous

I dont know how. I got the answer but i dont know how to get it??

- anonymous

That is why im asking .

- sasogeek

ok i'll try to actually solve it and post the answer, i just know there probably is something wrong with the solution above but i may be wrong. lemme try my hands on it and see what i get :) brb

- sasogeek

well at (1,3) the x=1 and y=3. we found the gradient of the tangent to be -1 and since it is a tangent to the curve with gradient function px^2-4x,
px^2-4x should be equal to -1 where x=1
hence
p(1)^2-4(1)=-1
p-4=-1
p=3

- anonymous

@sasogeek thanks! I get it. :]

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