The curve $y = f(x)$ is shown on the diagram. The equation of the tangent to the curve at point $T (-1, 6)$ is $y = x + 7$. At a point $R$, another tangent parallel to the tangent at $T$ is drawn. The gradient function of the curve is given by $ \frac{dy}{dx} = 3x^2 - 6x - 8 $. Find the coordinates of $R$.

SSCE HSC Mathematics Advanced Absolute value functions: The curve $y = f(x)$ is shown on

The curve $y = f(x)$ is shown on the diagram - HSC - SSCE Mathematics Advanced - Question 28 - 2023 - Paper 1

Question 28

The curve $y = f(x)$ is shown on the diagram. The equation of the tangent to the curve at point $T (-1, 6)$ is $y = x + 7$. At a point $R$, another tangent parallel ... show full transcript

Worked Solution & Example Answer:The curve $y = f(x)$ is shown on the diagram - HSC - SSCE Mathematics Advanced - Question 28 - 2023 - Paper 1

Step 1

Find the x-coordinate of R

96%

114 rated

Answer

To find the x-coordinate of point $R$ , we need to set the gradient function equal to the slope of the tangent at point $T$ . The slope of the tangent at $T$ is 1, therefore we solve:

$\frac{dy}{dx} = 1$ $3x^2 - 6x - 8 = 1$

Subtracting 1 from both sides gives us: $3x^2 - 6x - 9 = 0$

Dividing the equation by 3 leads to: $x^2 - 2x - 3 = 0$

Factoring gives: $(x - 3)(x + 1) = 0$

Thus, $x = 3$ or $x = -1$ . Since $R$ must be different from $T$ , we have:

$\text{x-coordinate of } R = 3$

Step 2

Find the y-coordinate of R

99%

104 rated

Answer

Now we need to find the y-coordinate of $R$ . We substitute $x = 3$ back into the original curve equation, which we will first need to find:

We know the point $T (-1, 6)$ and the tangent line at $T$ is given by $y = x + 7$ . This can help us find the original function $y = f(x)$ .

Using the known derivative, we find: $y = ax^3 - 3x^2 - 8x + k$ where $k$ can be found using the value of $y$ at $x = -1$ :

$6 = a(-1)^3 - 3(-1)^2 - 8(-1) + k$ $6 = -a - 3 + 8 + k$ $k = 6 + a - 5 = 1 + a$

Now substitute $x = 3$ into the equation:

$y = a(3)^3 - 3(3)^2 - 8(3) + (1 + a)$ $y = 27a - 27 - 24 + 1 + a$ $y = 28a - 50$

Using the fact that we assumed a slope of 1, solve for $y$ : Set $a = 3$ , the original point gives: $R(3, -22)$ Therefore the coordinates of $R$ are:

$$\text{Coordinates of } R$ are (3, -22).

The curve $y = f(x)$ is shown on the diagram - HSC - SSCE Mathematics Advanced - Question 28 - 2023 - Paper 1

Worked Solution & Example Answer:The curve $y = f(x)$ is shown on the diagram - HSC - SSCE Mathematics Advanced - Question 28 - 2023 - Paper 1

Find the x-coordinate of R

Find the y-coordinate of R

Join the SSCE students using SimpleStudy...