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maths911

  • 3 years ago

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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  1. sasogeek
    • 3 years ago
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    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 :)

  2. sasogeek
    • 3 years ago
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    |dw:1337323457233:dw|

  3. sasogeek
    • 3 years ago
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    umm the gradient function \(\ f'(x)=px^2-4x \) according to the question

  4. sasogeek
    • 3 years ago
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    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...

  5. maths911
    • 3 years ago
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    So If it is gradient function?It already refer to the graph?? Still didnt get it?

  6. sasogeek
    • 3 years ago
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    unless i'm wrong and mistaking, you might want to correct me then :/ idk

  7. maths911
    • 3 years ago
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    @sasogeek U r right. because the answer is p=3

  8. sasogeek
    • 3 years ago
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    how did you solve it maths911? just being curious

  9. maths911
    • 3 years ago
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    I dont know how. I got the answer but i dont know how to get it??

  10. maths911
    • 3 years ago
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    That is why im asking .

  11. sasogeek
    • 3 years ago
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    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

  12. sasogeek
    • 3 years ago
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    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

  13. maths911
    • 3 years ago
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    @sasogeek thanks! I get it. :]

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