- Thread starter QueekJJ
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Since you are lost in the beginning - let's start with definition:View attachment 25713

I had this question and got lost on where i should start. Also please do not give me answers straight away, just give me like ideas on what steps i should do so i can learn

What is the definition (mathematical) of the gradient of the curve?

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Here is a start: \(x^2+7x-9)=(2x+9)(x-1)\).View attachment 25713I had this question and got lost on where i should start. Also please do not give me answers straight away, just give me like ideas on what steps i should do so i can learn

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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

Just to confirm it, the answer gotten is 11 and -11. Am I correct?

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Answer for what?

Just to confirm it, the answer gotten is 11 and -11. Am I correct?

You were asked to calculate gradient !

Please share your work.

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?

\(\displaystyle \dfrac{dy}{dx} \text { is NOT equal to } 2x^2 + 7x - 4 \text { because } y = 2x^2 + 7x - 4.\)

What you could have shown is

\(\displaystyle f(x) = y = 2x^2 + 7x - 4.\)

\(\displaystyle y = 5 \implies 2x^2 + 7x - 4 = 5 \implies x = \dfrac{- 7 \pm \sqrt{49 - 4(2)(-9)}}{2 * 2} = \dfrac{-7 \pm {49 + 72}}{4} = 1 \text { or } - 4.5.\)

\(\displaystyle \dfrac{dy}{dx} = 4x + 7 = f’(x).\)

\(\displaystyle f’(1) = 4 * 1 + 7 = 11.\)

\(\displaystyle f’(-4.5) = 4(-4.5) + 7 = - 18 + 7 = - 11. \)

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 <3You wrote that 2x^2 + 7x - 4 = 4x+ 7. Do you know what '=' means?