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

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Question 28

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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

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Answer

To find the x-coordinate of point RR, we need to set the gradient function equal to the slope of the tangent line at point TT.

The slope of the tangent line at TT is 1, so we set:

dydx=1\frac{dy}{dx} = 1

Substituting the gradient function:

3x26x8=13x^2 - 6x - 8 = 1

Solving this gives:

3x26x9=0x22x3=03x^2 - 6x - 9 = 0 \Rightarrow x^2 - 2x - 3 = 0

Factoring the quadratic, we get:

(x3)(x+1)=0(x - 3)(x + 1) = 0

Thus, x=3x = 3 or x=1x = -1. Since point TT is at (1,6)(-1, 6), we discard 1-1 and take:

The x-coordinate of RR is 33.

Step 2

Find the y-coordinate of R

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Answer

Now we substitute x=3x = 3 back into the equation of the curve to find the y-coordinate of point RR. We will use the original gradient function:

y=x33x28x+ky = x^3 - 3x^2 - 8x + k

To find kk, we can substitute the known point T(1,6)T (-1, 6):

6=(1)33(1)28(1)+k6 = (-1)^3 - 3(-1)^2 - 8(-1) + k

This simplifies to:

6=13+8+k6=4+kk=26 = -1 - 3 + 8 + k \Rightarrow 6 = 4 + k \Rightarrow k = 2

So the equation of the curve is:

y=x33x28x+2y = x^3 - 3x^2 - 8x + 2

Now substituting x=3x = 3:

y=(3)33(3)28(3)+2y = (3)^3 - 3(3)^2 - 8(3) + 2

Calculating gives:

y=272724+2=22y = 27 - 27 - 24 + 2 = -22

Thus, the coordinates of RR are (3, -22).

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