How do you find the important points to graph #f(x)=x^2+2x#?

1 Answer
Leland Adriano Alejandro
Jun 11, 2018

Find the Vertex, X-intercepts, y-intercept

Explanation:

From the given #y=x^2+2x#

the format of #f(x)=y=ax^2+bx+c#

The Vertex #(h, k)#

with #a=1#and #b=2# and #c=0#

#h=-b/(2a)=-2/(2*1)=-1#

#k=c-b^2/(4a)=0-2^2/(4*1)=-4/4=-1#

Vertex #(h, k)=(-1, -1)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The x-intercepts #(x_1, 0)# and #(x_2, 0)#

From the given equation #y=x^2+2x#, set #y=0#,then solve for x values

#0=x^2+2x#
#0=x(x+2)#

then #x_1=0# and #x_2=-2#
The x-intercepts #(x_1, 0)# and #(x_2, 0)# are #(0, 0)#,and #(-2, 0)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The y-intercept #(0, y_1)#

From the given equation #y=x^2+2x#, set #x=0# then solve for the y value

#y=x^2+2x#

#y=0^2+2*(0)#

#y=0# when #x=0#

The y-intercept #(0, y_1)=(0, 0)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

enter image source here

I hope the explanation is useful....God bless

Related questions
See all questions in Graphs in the Coordinate Plane
Impact of this question
6881 views around the world
You can reuse this answer
Creative Commons License