# Question under differentiation chapter

Staff member

#### pka

##### Elite Member
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
Here is a start: $$x^2+7x-9)=(2x+9)(x-1)$$.

#### HallsofIvy

##### Elite Member
-"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.

• QueekJJ

#### apple2357

##### Full Member
• Jomo, QueekJJ and jonah2.0

#### QueekJJ

##### New member
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? #### Subhotosh Khan

##### Super Moderator
Staff member
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? Answer for what?

You were asked to calculate gradient !

Please share your work.

#### apple2357

##### Full Member
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?

#### JeffM

##### Elite Member
Your answer is correct, and I suspect your reasoning was correct, but your presentation is hard to follow and outright wrong in one respect.

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

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.

#### Jomo

##### Elite Member
You wrote that 2x^2 + 7x - 4 = 4x+ 7. Do you know what '=' means?

#### QueekJJ

##### New member
You wrote that 2x^2 + 7x - 4 = 4x+ 7. Do you know what '=' means?
Yeh i just realized my mistake on doing my working after what JeffM said. Thanks <3