math help (1 Viewer)

Dashdorm24

New Member

Need any help on these 2 questions, for an upcoming assessment.

cossine

Active Member
k = 1

you want to shift the curve 1 unit to the right. The graph P(x-k) is shifted k units to the right. So k = 1.

ii)
P(x) = (x+1)(x-2)^2

So just expand (x+1)(x-2)^2.

CM_Tutor

Moderator
Moderator
Actually, there are two possible values of $\bg_white k$...

Dashdorm24

New Member
Actually, there are two possible values of $\bg_white k$...
and what would that be ?

cossine

Active Member
and what would that be ?
I think he is referring case of k = -2 which I missed.

To shift a graph k units to the left the transformation is f(x+k) where k is positive.

CM_Tutor

Moderator
Moderator
I think he is referring case of k = -2 which I missed.

To shift a graph k units to the left the transformation is f(x+k) where k is positive.
Yes, I am.

And @Dashdorm24, you can check each is valid by doing the transformation and seeing that the resulting $\bg_white P(x)$ has $\bg_white x$ as a factor.

For example, taking $\bg_white P(x) = (x + 1)(x - 2)^2$ (which is the same as $\bg_white P(u) = (u + 1)(u - 2)^2$, I'm just using $\bg_white u$ as a dummy variable so that the transformation is clearer), and performing the transformation $\bg_white u = x - 1$, as suggested by @cossine, gives:

\bg_white \begin{align*} P(u) &= (u + 1)(u - 2)^2 \\ P(x - 1) &= \big[(x - 1) + 1\big]\big[(x - 1) - 2\big]^2 \quad \quad \text{putting u = x - 1} \\ &= x(x - 3)^2 \\ \text{So} \quad y &= x(x -3)^2 \quad \text{is the transformed polynomial} \end{align*}

and it has a single root at the origin and a double root at $\bg_white x = 3$, just as would be expected from a shift of one unit to the right.

The second transformation, $\bg_white u = x + 2$, which @cossine correctly identified from my hint, should produce a shift of 2 units to the left, moving the double root to the origin. Checking:

\bg_white \begin{align*} P(u) &= (u + 1)(u - 2)^2 \\ P(x + 2) &= \big[(x + 2) + 1\big)\big[(x + 2) - 2\big]^2 \quad \quad \text{putting u = x + 1} \\ &= (x + 3)x^2 \\ \text{So} \quad y &= x^2(x +3) \quad \text{is the transformed polynomial} \end{align*}

and it has a single root at $\bg_white x = -3$ and a double root at the origin, just as would be expected from a shift of two units to the left.

Noting that our given polynomial $\bg_white P(x) = (x + 1)(x - 2)^2$ has $\bg_white P(0) = 4$, we can also predict that a shift downwards by 4 should give a transformed polynomial with a root at the origin. Check:

\bg_white \begin{align*} Q(x) &= (x + 1)(x - 2)^2 \\ P(x) + 4 &= (x + 1)(x - 2)^2 \quad \quad \text{putting P(x) = Q(x) - 4} \\ P(x) &= (x + 1)(x - 2)^2 - 4 \\ &= (x + 1)\left(x^2 - 4x + 4\right) - 4 \\ &= x^3 - 4x^2 + 4x + x^2 - 4x + 4 - 4 \\ &= x^3 - 3x^2 \\ \text{So} \quad P(x) &= x^2(x - 3) \quad \text{is the transformed polynomial} \end{align*}

and it has a single root at $\bg_white x = 3$ and a double root at the origin, just as would be expected from a shift of four units to the down. (The point $\bg_white (3,\, 4)$ lies on the original polynomial, so a shift down of four should produce a root of the new polynomial at $\bg_white x = 3$.)

These techniques can be combined, along with dilations, to produce other polynomials as required.