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]}