equations (1 Viewer)

stressedadfff · Jul 18, 2021

how do you solve 10c? (also theres a typo its supposed to be x^3 for the first variable)

Screen Shot 2021-07-18 at 2.59.43 pm.png

quickoats · Jul 18, 2021

Consider the graph of $y = x^3 + x^2 -x + 2$ , most importantly, the stationery points. Also, consider the graph of y = k which should just be a horizontal line. There are 3 distinct solutions when these two graphs intersect each other 3 times.

stressedadfff · Jul 18, 2021

quickoats said:
Consider the graph of $y = x^3 + x^2 -x + 2$ , most importantly, the stationery points. Also, consider the graph of y = k which should just be a horizontal line. There are 3 distinct solutions when these two graphs intersect each other 3 times.

thanks, would you know the highlighted question please i dont get the 23:

Screen Shot 2021-07-18 at 3.24.57 pm.png

Screen Shot 2021-07-18 at 3.28.35 pm.png

jimmysmith560 · Jul 18, 2021

Question 15c appeared in a 2000 "Mathematics 2/3 Unit (Common)" paper. Here is what was suggested from the 2000 Examination Report:

Sorry that I'm unable to provide more direct help, but hopefully this feedback can perhaps guide your thinking?

CM_Tutor · Jul 18, 2021

stressedadfff said:
View attachment 31159

$\begin{align*} y &= x^4 - 4x^3 + 7 \\ \cfrac{dy}{dx} &= 4x^3 - 12x^2 \\ \text{At $(2,\,-9)$:} \quad m_{\text{tang}} &= 4(2)^3 - 12(2)^2 \\ &= 32 - 48 \\ &= -16 \\ \text{We can find the tangent using the point-gradient formula} \quad y - y_1 &= m(x - x_1) \quad \text{where $(x_1,\,y_1)$ is the point on the line and $m$ is its gradient (here, $m=m_{\text{tang}}$).} \\ \text{So, the tangent at $(2,\,-9)$ is} \quad y - (-9) &= -16(x - 2) \\ y + 9 &= -16x + 32 \\ 16x + y + 9 - 32 &= 0 \\ 16x + y + - 23 &= 0 \quad \text{as required} \end{align*}$

Eagle Mum · Jul 18, 2021

CM_Tutor said:
$\begin{align*} y &= x^4 - 4x^3 + 7 \\ \cfrac{dy}{dx} &= 4x^3 - 12x^2 \\ \text{At $(2,\,-9)$:} \quad m_{\text{tang}} &= 4(2)^3 - 12(2)^2 \\ &= 32 - 28 \\ &= -16 \\ \text{We can find the tangent using the point-gradient formula} \quad y - y_1 &= m(x - x_1) \quad \text{where $(x_1,\,y_1)$ is the point on the line and $m$ is its gradient (here, $m=m_{\text{tang}}$).} \\ \text{So, the tangent at $(2,\,-9)$ is} \quad y - (-9) &= -16(x - 2) \\ y + 9 &= -16x + 32 \\ 16x + y + 9 - 32 &= 0 \\ 16x + y + - 23 &= 0 \quad \text{as required} \end{align*}$

There’s a trivial typo in line 4: 32 - 48 = -16

CM_Tutor · Jul 18, 2021

Eagle Mum said:
There’s a trivial typo in line 4: 32 - 48 = -16

You are correct, and thanks for pointing this out. Typo corrected.

stressedadfff · Jul 19, 2021

thanks!!!

equations (1 Viewer)

stressedadfff

Well-Known Member

quickoats

Well-Known Member

stressedadfff

Well-Known Member

jimmysmith560

Le Phénix Trilingue

CM_Tutor

Moderator

Eagle Mum

Well-Known Member

CM_Tutor

Moderator

stressedadfff

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)