How do you graph the quadratic function and identify the vertex and axis of symmetry and x intercepts for #y=(x-2)(x-6)#?

2 Answers
Feb 17, 2018

Please follow from the explanation.

Explanation:

To find the vertex (commonly known as the turning or stationary point), we can employ several approaches. I will employ calculus to do this.

First Approach:
Find the derivative of the function.

Let #f(x)=y = (x-2)(x-6)#
then, #f(x)= x^2-8x+12#

the derivative of the function (using the power rule) is given as
#f'(x)=2x-8#
We know that the derivative is naught at the vertex. So,
#2x-8=0#
#2x=8#
#x=4#
This gives us the x-value of the turning point or vertex. We will now substitute #x=4# into #f# to obtain the corresponding y-value of the vertex.
that is, #f(4)=(4)^2-8(4)+12#
#f(4)=-4#
Hence the co-ordinates of the vertex are #(4,-4)#

Any quadratic function is symmetrical about the line running vertically through its vertex.. As such, we have found the axis of symmetry when we found the co-ordinates of the vertex.
That is, the axis of symmetry is #x=4#.

To find x-intercepts: we know that the function intercepts the x-axis when #y=0#. That is, to find the x-intercepts we have to let #y=0#.
#0=(x-2)(x-6)#
#x-2=0 or x-6=0#
therefore, #x=2 or x=6 #
This tells us that the co-ordinates of the x-intercept are #(2,0)# and #(6,0)#

To find the y-intercept, let #x=0#
#y=(0-2)(0-6)#
#y=12#
This tells us that the co-ordinate of the y-intercept is #0,12#
Now use the points we derived above to graph the function graph{x^2 - 8x +12 [-10, 10, -5, 5]}

Feb 17, 2018

#"see explanation"#

Explanation:

#"to find the intercepts"#

#• " let x = 0, in the equation for y-intercept"#

#• " let y = 0, in the equation for x-intercepts"#

#x=0toy=(-2)(-6)=12larrcolor(red)"y-intercept"#

#y=0to(x-2)(x-6)=0#

#"equate each factor to zero and solve for x"#

#x-2=0rArrx=2#

#x-6=0rArrx=6#

#rArrx=2,x=6larrcolor(red)"x-intercepts"#

#"the axis of symmetry goes through the midpoint"#
#"of the x-intercepts"#

#x=(2+6)/2=4rArrx=4larrcolor(red)"axis of symmetry"#

#"the vertex lies on the axis of symmetry, thus has"#
#"x-coordinate of 4"#

#"to obtain y-coordinate substitute "x=4" into the"#
#"equation"#

#y=(2)(-2)=-4#

#rArrcolor(magenta)"vertex "=(4,-4)#

#"to determine if vertex is max/min consider the"#
#"value of the coefficient a of the "x^2" term"#

#• " if "a>0" then minimum"#

#• " if "a<0" then maximum"#

#y=(x-2)(x-6)=x^2-8x+12#

#"here "a>0" hence minimum "uuu#

#"gathering the information above allows a sketch of "#
#"quadratic to be drawn"#
graph{(y-x^2+8x-12)(y-1000x+4000)=0 [-10, 10, -5, 5]}