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.

1. 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 :)

2. sasogeek

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3. sasogeek

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

4. 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...

5. maths911

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

6. sasogeek

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

7. maths911

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

8. sasogeek

how did you solve it maths911? just being curious

9. maths911

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

10. maths911

That is why im asking .

11. 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

12. 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

13. maths911

@sasogeek thanks! I get it. :]