How do you find the axis of symmetry, vertex and x intercepts for #y=x^2+2x#?

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Answer 1 · 2017-11-05T13:58:30

Axis of Symmetry : #x = -1#

Vertex = ( - 1, - 1 )

X-Intercept = There are two values : ( - 2 ) and Zero

We are given the Quadratic #y = x^2 + 2 * x#

We will now find the Vertex Form.

In general, Vertex Form is written as

#y = f( x ) = a*(x - h) ^ 2 + k#

Our Vertex will be ( h, k )

#f( x ) = [ x ^ 2+ 2 * x + ( ) ] - ( )#

(Divide the coefficient of the x-term by 2 and square it: (2/2) ^ 2, to add and subtract)

#f( x ) = [ x ^ 2+ 2 * x + ( 1 ) ] - ( 1 )#

We can write the above as

#f( x ) = ( x + 1) ^ 2 - 1 #

Now, #h = (-1)# and #k = (- 1)#, in the general Vertex Form.

We now have the Vertex Form of the Quadratic

In our problem situation, our Vertex is ( - 1, - 1 )

Axis of Symmetry is given by: x = h

Hence, x = ( - 1 ) is the required Axis of Symmetry

To find the x-Intercept, equate f( x ) = 0

#0 = (x + 1) ^ 2 - 1#

# (x + 1) ^ 2 - 1 = 0#

# (x + 1) ^ 2 = 1#

#(x + 1) = +- sqrt(1)#

This gives us ( - 2) and Zero as the two x-Intercepts.