# 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

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

then,

the derivative of the function (using the power rule) is given as

We know that the derivative is naught at the vertex. So,

This gives us the x-value of the turning point or vertex. We will now substitute

that is,

Hence the co-ordinates of the vertex are

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

To find x-intercepts: we know that the function intercepts the x-axis when

therefore,

This tells us that the co-ordinates of the x-intercept are

To find the y-intercept, let

This tells us that the co-ordinate of the y-intercept is

Now use the points we derived above to graph the function graph{x^2 - 8x +12 [-10, 10, -5, 5]}

#### 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]}