<p>A function is defined as&nbsp;$\mathrm{f}\left( x \right)=2{{x}^{3}}-6{{x}^{2}}+6x-12$.</p>
<ol class="i-list"><li data-list-item-id="item-1759131883284-lvvdjugk8" data-list-item-path="item-1759131883284-lvvdjugk8"><div class="list-item-paragraph">Show that&nbsp;$\mathrm{f}\left( x \right)$ can be written in the form $p\left\{ {{\left( x+q \right)}^{3}}+r \right\}$, where&nbsp;$p,q$ and $r$ are constants to be found.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1759131884949-ca1o9su93" data-list-item-path="item-1759131884949-ca1o9su93"><div class="list-item-paragraph">Hence, or otherwise, describe a sequence of transformations that transform the graph of $y={{x}^{3}}$ onto the graph of $y=\mathrm{f}\left( x \right)$.<span class="inline-marks">[3]</span></div></li></ol>

A function is defined as $\mathrm{f}\left( x \right)=2{{x}^{3}}-6{{x}^{2}}+6x-12$.

Show that $\mathrm{f}\left( x \right)$ can be written in the form $p\left\{ {{\left( x+q \right)}^{3}}+r \right\}$, where $p,q$ and $r$ are constants to be found.

Hence, or otherwise, describe a sequence of transformations that transform the graph of $y={{x}^{3}}$ onto the graph of $y=\mathrm{f}\left( x \right)$.

Default solution

Comparing the coefficient of $x^3$ gives $p=2$.

$\displaystyle f(x)=2x^3-6x^2+6x-12=2[(x+q)^3+r]$.

Expanding, $2[(x+q)^3+r]=2x^3+6qx^2+6q^2x+2q^3+2r$.

Comparing coefficients: $6q=-6$, so $q=-1$; $6q^2=6$, consistent; and $2q^3+2r=-12$, so $-2+2r=-12$, giving $r=-5$.

Hence $f(x)=2[(x-1)^3-5]$, with $q=-1$, $p=2$, and $r=-5$.

Starting from $y=x^3$, first replace $x$ by $x-1$, translating 1 unit in the positive $x$-direction.

Then replace $y$ by $y+5$, translating 5 units in the negative $y$-direction.

Finally replace $y$ by $\frac y2$, scaling by factor 2 parallel to the $y$-axis.

N2019 P1 Q3

N2019 P1 Q3

Video Solution:

Default solution