Question under differentiation chapter
- Thread starter QueekJJ
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D
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Since you are lost in the beginning - let's start with definition:
What is the definition (mathematical) of the gradient of the curve?
pka
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Here is a start: \(x^2+7x-9)=(2x+9)(x-1)\).
HallsofIvy
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-"The curve \(\displaystyle y= 2x^2+ 7x- 4\) and the line y= 5 meet at the points P and Q."
The first thing you need to do is find the points P and Q: solve \(\displaystyle 2x^2+ 7x- 4= 5\).
That is the same as \(\displaystyle 2x^2+ 7x- 9= 0\) and, as pka points out, \(\displaystyle 2x^2+ 7X- 9= (2x+ 9)(x- 1)\). Once you know P and Q, evaluate the derivative of \(\displaystyle 2x^2+ 7x- 9= 0\) at those points.
The first thing you need to do is find the points P and Q: solve \(\displaystyle 2x^2+ 7x- 4= 5\).
That is the same as \(\displaystyle 2x^2+ 7x- 9= 0\) and, as pka points out, \(\displaystyle 2x^2+ 7X- 9= (2x+ 9)(x- 1)\). Once you know P and Q, evaluate the derivative of \(\displaystyle 2x^2+ 7x- 9= 0\) at those points.
Ah sorry for the late reply, wasn't able to use my gadget the pass few days... Thanks apple2357 and Hallsoflvy for the help, i now finally understood the question and the steps to answer it.
Just to confirm it, the answer gotten is 11 and -11. Am I correct?
Just to confirm it, the answer gotten is 11 and -11. Am I correct?
D
Deleted member 4993
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Answer for what?Ah sorry for the late reply, wasn't able to use my gadget the pass few days... Thanks apple2357 and Hallsoflvy for the help, i now finally understood the question and the steps to answer it.
Just to confirm it, the answer gotten is 11 and -11. Am I correct?
You were asked to calculate gradient !
Please share your work.
Ah sorry for the late reply, wasn't able to use my gadget the pass few days... Thanks apple2357 and Hallsoflvy for the help, i now finally understood the question and the steps to answer it.
Just to confirm it, the answer gotten is 11 and -11. Am I correct?
I can confirm the gradient at the two points of intersection are 11 and -11.
Can you see how the graph suggests a symmetry?
Your answer is correct, and I suspect your reasoning was correct, but your presentation is hard to follow and outright wrong in one respect.
[MATH]\dfrac{dy}{dx} \text { is NOT equal to } 2x^2 + 7x - 4 \text { because } y = 2x^2 + 7x - 4.[/MATH]
What you could have shown is
[MATH]f(x) = y = 2x^2 + 7x - 4.[/MATH]
[MATH]y = 5 \implies 2x^2 + 7x - 4 = 5 \implies x = \dfrac{- 7 \pm \sqrt{49 - 4(2)(-9)}}{2 * 2} = \dfrac{-7 \pm {49 + 72}}{4} = 1 \text { or } - 4.5.[/MATH]
[MATH]\dfrac{dy}{dx} = 4x + 7 = f’(x).[/MATH]
[MATH]f’(1) = 4 * 1 + 7 = 11.[/MATH]
[MATH]f’(-4.5) = 4(-4.5) + 7 = - 18 + 7 = - 11. [/MATH]
Good work. Correct answers are a whole lot more important than presentation that is sweet and neat.
[MATH]\dfrac{dy}{dx} \text { is NOT equal to } 2x^2 + 7x - 4 \text { because } y = 2x^2 + 7x - 4.[/MATH]
What you could have shown is
[MATH]f(x) = y = 2x^2 + 7x - 4.[/MATH]
[MATH]y = 5 \implies 2x^2 + 7x - 4 = 5 \implies x = \dfrac{- 7 \pm \sqrt{49 - 4(2)(-9)}}{2 * 2} = \dfrac{-7 \pm {49 + 72}}{4} = 1 \text { or } - 4.5.[/MATH]
[MATH]\dfrac{dy}{dx} = 4x + 7 = f’(x).[/MATH]
[MATH]f’(1) = 4 * 1 + 7 = 11.[/MATH]
[MATH]f’(-4.5) = 4(-4.5) + 7 = - 18 + 7 = - 11. [/MATH]
Good work. Correct answers are a whole lot more important than presentation that is sweet and neat.
Yeh i just realized my mistake on doing my working after what JeffM said. Thanks <3