Question

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

Answer 1

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]}

Answer 2

`"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]}