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)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

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