How do you sketch the graph of y=x^2+8y=x2+8 and describe the transformation?

2 Answers
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+8x2+8

In this case x should equal 00

so y=0^2+8y=02+8

giving just 8

the y-intercept is 8 as c=8c=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^2x2 but raised by 8 on the y-axis.

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

Axis of symmetry -> y-axis

y-intercept = 8

x-intercept -> none

Explanation:

color(blue)("General shape")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^2x2 term is positive so the general form is that of uu
,..................................................................................................

color(blue)("Axis of symmetry")Axis of symmetry

Consider the standardised equation form of y=ax^2+bx+cy=ax2+bx+c

Write this as: y=a(x^2+b/a x)+cy=a(x2+bax)+c then we have we can use the rather nifty trick of:

x_("vertex")=(-1)xx b/(2a)xvertex=(1)×b2a

or the same thing in a different form:

x_("vertex")=(-1/2)xx b/(a)xvertex=(12)×ba

Note that in the case of this question

a=1 ->ax^2->1xx x^2 = x^2a=1ax21×x2=x2

So b/a->b/1=bbab1=b

Now compare ax^2+bx+c color(white)(b)" to " color(white)(b)x^2+8ax2+bx+cb to bx2+8

x^2+8x2+8 does not have a bxbx term. To make this happen b=0b=0

so x_("vertex") = (-1/2)xxb/a -> (-1/2)xx0/1 = 0xvertex=(12)×ba(12)×01=0

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

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

color(blue)("Determine the y intercept and x intercepts")Determine the y intercept and x intercepts

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

y=x^2+8" "->" "y=0+8y=x2+8 y=0+8

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

As the vertex is at (0,8)(0,8) and the graph is of form uu 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 uu central about the y-axis and has the vertex passing through the labeled point (x,y)=(0,8)

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.

Tony B