This is a step by step solution for Q5 in the Extended Response Section of VCAA Exam 2 , 2014.

You should try to solve the problem yourself, and only look at the slide where you are up to if you cannot make any further progress.

When you view a slide, use the hint on it to try to keep going yourself.

In other words, only look at a slide as the last resort.

GOOD LUCK!

Exam 2 2014 QUESTION 5.pdf

First take out the HCF: and recall that we can factorize the difference of two cubes!!

OR: simply factorize on CAS:

CAS output is )

a.

Now look at the format that they have asked for.They are asking for the quadratic factor to be in turning point form!

Complete the square.

b.We have found that:

We are given that:

If we compare the quadratic factors of both functions, we see that the quadratic component of g(x) could be obtained from that of f(x) by replacing x with x + 1.

Test out this idea!

Since the graph of is obtained by translating f(x) by 1 unit in the negative x-direction.

If the graph of f (blue) is shifted 1 unit to the left, there is one positive x-intercept. If shifted 2.4 units to the left, there is one positive x-intercept.

c.i

If shifted 3 units to the left, there are no more positive x-intercepts. Likewise if shifted any further to the left, there will be no positive x-intercepts.

So the values of d for which there is ONE positive x-intercept are:

OR you can write it as

d.

To answer this question once again a graph is essential. We wish to find the value of n for which the horizontal line will cut the graph in only one point.

Try this if you haven’t yet done so, and see if you can solve the problem yourself.

In this position, with n = - 11.7, there is no solution for the equation:

We need to find the exact y value of the turning point of g(x). This will be the value of n we require.

For a stationary point,

Now evaluate: )

Required value of n is:

e.

In this question, we are told that:

where m is a positive number.In other words, we are being told that

…and we are being asked for an equation connecting u and v. See if you can go ahead and solve the problem now.

We know already:

Therefore:

So:

e ii

We are told that:

From the previous question, we have found that:

Solving the simultaneous equations on CAS:

and or and

To decide which values are appropriate we must refer again to the graph of g(x).

We see from the graph that since the gradient at the point where x = u is positive(equal to m), and the gradient at x = v is negative (equal to –m), it follows that u > v.

and

We conclude:

f.i

To find the equation of the tangent at it means that p is the parameter, and we must find the gradient and the y-intercept of the line in terms of p.

First, we find the gradient of the tangent, which is equal to

The point ( is the point:

If you haven’t done this yet, find the equation now!!

Using:

We get:

𝒚=(𝟒𝒑𝟑−𝟖 )𝒙 −𝒑 (𝟒𝒑𝟑−𝟖 )+𝒑𝟒−𝟖𝒑

𝒚=(𝟒𝒑𝟑−𝟖 )𝒙 −𝟑𝒑𝟒

f.Ii

We require that the tangent found above goes through the point

We substitute these co-ordinates into the tangent equation!

Do it now if you haven’t yet!!

12 = (4×

Required values of p are p = 0 or p = 2Now find the equations of the tangent lines that we require:

𝒚=(𝟒𝒑𝟑−𝟖 )𝒙 −𝟑𝒑𝟒

If

If

𝒚=𝟐𝟒𝒙−𝟒𝟖

The two tangent equations are: