# How do you sketch the graph of y=x^2-5 and describe the transformation?

May 10, 2018

See below

#### Explanation:

If you know the graph of a function $y = f \left(x\right)$, then you can have four kind of transformations: the most general expression is

$A f \left(w x + h\right) + v$

where:

• $A$ multiplies the whole function, thus stretching it vertically (expansion if if $| A | > 1$, contraction otherwise)
• $w$ multiplies the input variable, thus stretching it horizontally (expansion if if $| w | < 1$, contraction otherwise)
• $h$ and $v$ are, respectively, horizontal and vertical translations.

In your case, starting from $f \left(x\right) = {x}^{2}$, you have $A = 1$ and $w = 1$. Being multiplicative factors, they have non effect.

Moreover, $h = 0$. Being ad additive factor, it has non effect.

Finally, you have $v = - 5$. This means that, if you start from the "standard" parabola $f \left(x\right) = {x}^{2}$, the graph of $f \left(x\right) = {x}^{2} - 5$ is the same, just translated $5$ units down.