# How do you sketch the graph of y=0.5(x-2)^2-2 and describe the transformation?

Aug 10, 2017
• The y-intercept is at the origin, $\left(0 , 0\right)$.
• The x-intercepts are
$\left(0 , 0\right)$ and $\left(4 , 0\right)$.
• The turning point is $\left(2 , - 2\right)$.
graph{ y = 0.5(x-2)^2-2 [-10, 10, -5, 5]}

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 $\frac{1}{2}$.

#### Explanation:

Since the equation is in the form of $y = a {\left(x - h\right)}^{2} + k$, 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, let $x - 0$.
• To find the x-intercepts, let $y = 0$, factorise the equation, and solve for $x$.

Let $x = 0$:
$y = 0.5 {\left(\left(0\right) - 2\right)}^{2} - 2$
$y = 0$

The y-intercept is at the origin, $\left(0 , 0\right)$.

Let $y = 0$:

$0 = 0.5 {\left(x - 2\right)}^{2} - 2$

$0 = 0.5 \left({\left(x - 2\right)}^{2} - 4\right)$

$0 = {\left(x - 2\right)}^{2} - 4$

$0 = \left(\left(x - 2\right) + 2\right) \left(\left(x - 2\right) - 2\right)$

$0 = x \left(x - 4\right)$

$x - 4 = 0$, $x = 0$
$x = 4$, $x = 0$

Therefore, x-intercepts are $\left(0 , 0\right)$ and $\left(4 , 0\right)$.

Now that you have all the information, let's graph it along with y=x^2.

graph{y=0.5(x-2)^2 - 2 [-10, 10, -5, 5]}

As you can see, 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 $\frac{1}{2}$.