Question

How do you sketch the graph of $y=0.5{{\left( x-2 \right)}^{2}}-2$ and describe the transformation?

Answer 1

In order to find the solution of the given question that is to find how to sketch the graph of $y=0.5{{\left( x-2 \right)}^{2}}-2$ and describe the transformation, determine the form of the given equation with the help of the standard formula that is $y=a{{\left( x-h \right)}^{2}}+k$ then determine the turning point that is find $\left( h,k \right)$ by comparing the given equation with the standard formula that is $y=a{{\left( x-h \right)}^{2}}+k$. Now find the $y$-intercept by putting $x=0$, factorise the equation and solve for $y$and also find the $x$-intercept by putting $y=0$, factorise the equation and solve for $x$. After getting these points you’ll be able to sketch the graph and describe the transformation.

Complete step by step solution:
According to the question, given equation in the question is as follows:
$y=0.5{{\left( x-2 \right)}^{2}}-2...\left( 1 \right)$
As we can see the form of the above equation is parabola means it’s in the form of the standard formula that is $y=a{{\left( x-h \right)}^{2}}+k$.
To determine the turning point that is find $\left( h,k \right)$, compare the given equation with the standard formula, we get that
$\Rightarrow h=2$ and $k=-2$
With this we can interpret that the graph's turning point shifts $2$ units to the right, and $2$units down.
Therefore, the turning point is $\left( 2,-2 \right)$.
The next step is to find the intercepts. Recall that to find the $y$-intercept by putting $x=0$, we get:
$\Rightarrow y=0.5{{\left( \left( 0 \right)-2 \right)}^{2}}-2$
Now factorise the above equation and solve for $y$, we get:
$\Rightarrow y=0.5{{\left( -2 \right)}^{2}}-2$
$\Rightarrow y=0.5\times 4-2$
$\Rightarrow y=2-2$
$\Rightarrow y=0$
The $y$-intercept is at the origin, $\left( 0,0 \right)$.
Now to find the $x$-intercept by put $y=0$ in equation $\left( 1 \right)$, we get:
$\Rightarrow 0=0.5{{\left( x-2 \right)}^{2}}-2$
To simplify it further factorise the above equation and solve for $x$, we get:
$\Rightarrow 0.5{{\left( x-2 \right)}^{2}}=2$
$\Rightarrow {{\left( x-2 \right)}^{2}}=\dfrac{2}{0.5}$
$\Rightarrow {{\left( x-2 \right)}^{2}}=4$
$\Rightarrow {{\left( x-2 \right)}^{2}}-4=0$
$\Rightarrow {{\left( x-2 \right)}^{2}}-{{2}^{2}}=0$
Now to simplify it further apply the identity $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$ where $a=\left( x-2 \right)$ and $b=2$in the above equation, we get:
$\Rightarrow \left( \left( x-2 \right)+2 \right)\left( \left( x-2 \right)-2 \right)=0$
$\Rightarrow x\left( x-4 \right)=0$
$\Rightarrow x=0$ and $\left( x-4 \right)=0$
$\Rightarrow x=0,4$
Therefore, $x$-intercepts are $\left( 0,0 \right)$and $\left( 4,0 \right)$.
Now as we have all the information, we can sketch the graph as shown in the figure below:

As you can see, the transformation in the above graph, that is the graph has shifted $2$ units to the right and $2$ units down compared to $y={{x}^{2}}$. The graph is also wider than $y={{x}^{2}}$. Since the value of $a$ in the equation is $\dfrac{1}{2}$.
Whereas the $y$-intercept is at the origin, $\left( 0,0 \right)$and the $x$-intercepts are $\left( 0,0 \right)$and $\left( 4,0 \right)$.Also, The turning point is $\left( 2,-2 \right)$.

Note: Students can go wrong while calculating the $y$-intercept and the $x$-intercepts. They make mistakes by letting $x=0$ to find the $x$-intercept and let $y=0$ to find the $y$-intercept which is completely wrong and further leads to the no answer. It’s important to remember to find the $y$-intercept, put $x=0$and solve for $y$. To find the $x$-intercept, put $y=0$ and solve for $x$.