Problems involving integration: quadratic's gradient at (-2,3) is -1

[almas]

Hi,

Any help would be greatly appreciated. I am studying A-level maths from home myself from books and free internet resources. I came across a question which I'm stuck on. A quadratic curve passes through O(0,0) and its gradient at the pint (-2,3) is -1. Find the equation of the curve.

I have done other questions from the same section in the book but managed them because they gave me the differential so I integrated it and got the answer. As no differential has been given I am at a loss with this and can't seem to figure out where I should start.

Thank you in advance for your help, it's great that you are helping students for free. I have been stuck on the question for a while so would be indebted with your help.

Best wishes

 

[Deleted member 4993]

Hi,

Any help would be greatly appreciated. I am studying A-level maths from home myself from books and free internet resources. I came across a question which I'm stuck on. A quadratic curve passes through O(0,0) and its gradient at the pint (-2,3) is -1. Find the equation of the curve.

I have done other questions from the same section in the book but managed them because they gave me the differential so I integrated it and got the answer. As no differential has been given I am at a loss with this and can't seem to figure out where I should start.

Thank you in advance for your help, it's great that you are helping students for free. I have been stuck on the question for a while so would be indebted with your help.

Best wishes

What is the general equation of a quadratic curve? → y = Ax2 + Bx + C

What is the general equation of a quadratic curve - that passes through the origin?

 

[almas]

What is the general equation of a quadratic curve? → y = Ax2 + Bx + C

What is the general equation of a quadratic curve - that passes through the origin?

Hi, thanks for the response, the question just says "in each of these find the equation of the curve ..(c) A quadratic passes through O (0,0) and it's gradient at the point (-2,3) is -1". That is the full question without any further information or the general equation of the curve.

Thanks

 

[ksdhart2]

Right, the problem text doesn't give the answer to Subhotosh Khan's question. That's information you're expected to give, based on the information in the problem text You know the general equation of a quadratic (or if you didn't know before, you do now), so think about what it means for that most general quadratic to pass through the point (0,0). It means that when x = 0, y (aka f(x)) is also 0. How does that change the equation of the quadratic? What information can you gather about any of the coefficients A, B, or C based on that? You're then given information about the gradient (the "slope" of the curve, aka the derivative) at the point (-2,3). That implies that the quadratic must also pass through that point. How does that change the equation? Now, if the derivative of the curve at the point (-2,3) is -1, what does that mean? What is the derivative of f(x)=Ax^2 + Bx + C? What does it mean that it has a value of -1 at the point (-2,3)? All of this gives you two equations, so you can solve that system using whatever method you prefer.

 

[HallsofIvy]

Are you sure you have copied the problem correctly? You say, in the title, that it "involves integration" but I see no integration involved. Further, since the "general quadratic function", \(\displaystyle y= ax^2+ bx+ c\), involves three unknown coefficients, it requires three equations, so three conditions, to find a, b, and c and you only give 2.

From the information that y(0)= 0 we get \(\displaystyle a(0)^2+ b(0)+ c= c= 0\). From the information that y'(-2)= -1, since \(\displaystyle y'= 2ax+ b\), we get \(\displaystyle 2a(-2)+ b= -4a+ b= -1\) so that \(\displaystyle b= 4a- 1\) but then we have no way to determine "a". Any quadratic function of the form \(\displaystyle y= ax^2+ (4a- 1)x\) satisfies the conditions of this problem for any value of a.

 

[Deleted member 4993]

Are you sure you have copied the problem correctly? You say, in the title, that it "involves integration" but I see no integration involved. Further, since the "general quadratic function", \(\displaystyle y= ax^2+ bx+ c\), involves three unknown coefficients, it requires three equations, so three conditions, to find a, b, and c and you only give 2.

From the information that y(0)= 0 we get \(\displaystyle a(0)^2+ b(0)+ c= c= 0\). From the information that y'(-2)= -1, since \(\displaystyle y'= 2ax+ b\), we get \(\displaystyle 2a(-2)+ b= -4a+ b= -1\) so that \(\displaystyle b= 4a- 1\) but then we have no way to determine "a". Any quadratic function of the form \(\displaystyle y= ax^2+ (4a- 1)x\) satisfies the conditions of this problem for any value of a.

The other equation is embedded in the fact that (-2,1) is a point on the curve.