# How do you sketch the graph of y=x^2+8 and describe the transformation?

Aug 22, 2017

See below

#### Explanation:

The first thing you need to find is the axis of symmetry by using

-b/2a

after you find that, use it to find the vertex of ${x}^{2} + 8$

In this case x should equal $0$

so $y = {0}^{2} + 8$

giving just 8

the y-intercept is 8 as $c = 8$

the graph should look like this

graph{x^2+8 [-11.5, 8.5, 0.84, 10.84]}

it is a parabola.

Aug 22, 2017

Detailed explanation given

The transformation is that of ${x}^{2}$ but raised by 8 on the y-axis.

Vertex $\to \left(x , y\right) = \left(0 , + 8\right)$

Axis of symmetry $\to$ y-axis

y-intercept = 8

x-intercept -> none

#### Explanation:

$\textcolor{b l u e}{\text{General shape}}$

The first thing you need to determine is the general shape.
This is a quadratic equation so it has a horse shoe type shape.

The ${x}^{2}$ term is positive so the general form is that of $\cup$
,..................................................................................................

$\textcolor{b l u e}{\text{Axis of symmetry}}$

Consider the standardised equation form of $y = a {x}^{2} + b x + c$

Write this as: $y = a \left({x}^{2} + \frac{b}{a} x\right) + c$ then we have we can use the rather nifty trick of:

${x}_{\text{vertex}} = \left(- 1\right) \times \frac{b}{2 a}$

or the same thing in a different form:

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a}$

Note that in the case of this question

$a = 1 \to a {x}^{2} \to 1 \times {x}^{2} = {x}^{2}$

So $\frac{b}{a} \to \frac{b}{1} = b$

Now compare $a {x}^{2} + b x + c \textcolor{w h i t e}{b} \text{ to } \textcolor{w h i t e}{b} {x}^{2} + 8$

${x}^{2} + 8$ does not have a $b x$ term. To make this happen $b = 0$

so ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a} \to \left(- \frac{1}{2}\right) \times \frac{0}{1} = 0$

Thus the vertex coincides with the y-axis as does the axis of symmetry .

'.........................................................................

$\textcolor{b l u e}{\text{Determine the y intercept and x intercepts}}$

As the axis of symmetry is the y-axis set $x = 0$ giving:

$y = {x}^{2} + 8 \text{ "->" } y = 0 + 8$

y_("intercept")->"Vertex "->(x,y)=(0,+8)

As the vertex is at $\left(0 , 8\right)$ and the graph is of form $\cup$ the x-intercepts do not exist.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
color(blue)("Drawing the "ul("sketch"))

You do not need to work out a whole pile of points. Just make sure the general shape is $\cup$ central about the y-axis and has the vertex passing through the labeled point $\left(x , y\right) = \left(0 , 8\right)$

You do not need to draw to scale so it should only take about 5 to 15 seconds to complete ( some may take longer ). Do not forget to label your points and put a title on it.