Consider f(x) = px^2+qx, p, q constants, (1, 3) lying on the curve. Tangent at (1, 3) has gradient 8. Find p, q.

[asmdmasd]

Hello,

I need help with this problem.

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Consider the curve with equation [math]f(x)\, =\, px^2\, +\, qx,[/math] where p and q are constants. The point A(1, 3) lies on the curve. The tangent to the curve at A has gradient 8. Find the value of p and of q.

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I think my misunderstanding stems from what to do with p and q when you differentiate f(x). Thank you.

 

[lev888]

Hello,

I need help with this problem.

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Consider the curve with equation [math]f(x)\, =\, px^2\, +\, qx,[/math] where p and q are constants. The point A(1, 3) lies on the curve. The tangent to the curve at A has gradient 8. Find the value of p and of q.

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I think my misunderstanding stems from what to do with p and q when you differentiate f(x). Thank you.

Try substituting numbers, see if there is a pattern.

 

[topsquark]

Hello,

I need help with this problem.

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Consider the curve with equation [math]f(x)\, =\, px^2\, +\, qx,[/math] where p and q are constants. The point A(1, 3) lies on the curve. The tangent to the curve at A has gradient 8. Find the value of p and of q.

---

I think my misunderstanding stems from what to do with p and q when you differentiate f(x). Thank you.

You know that A is on the curve so you immediately have the equation
\(\displaystyle 3 = p(1)^2 + q(1)\)

You also know that the tangent to the curve at A is 8. So
\(\displaystyle \dfrac{df}{dx} = 8\)
at x = 1.

Can you finish?

-Dan

 

[Dr.Peterson]

Hello,

I need help with this problem.

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Consider the curve with equation [math]f(x)\, =\, px^2\, +\, qx,[/math] where p and q are constants. The point A(1, 3) lies on the curve. The tangent to the curve at A has gradient 8. Find the value of p and of q.

---

I think my misunderstanding stems from what to do with p and q when you differentiate f(x). Thank you.

Since p and q are constants, they behave just like numbers. Do the same thing you would do to differentiate f(x) = 3x^2 + 5x, for example.

 

[HallsofIvy]

You probably know the basic derivative law (cf(x))'= cf'(x) for any constant, c. It doesn't matter whether that constant is called "c" or "p" or "q"! The derivative of \(\displaystyle px^2\) is \(\displaystyle p(2x)= 2px\) and the derivative of \(\displaystyle qx\) is \(\displaystyle q(1)= q\).