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January 28, 2015

January 28, 2015

Posted by **Jimmy-can someone please help** on Tuesday, December 18, 2012 at 9:40pm.

a. zeros and the multiplicity of each

b. number of turning points

c. end behavior

f(x)=(x-1)(x-5)

a. x=1 and x=5--is this correct?

b. number of turning points

I had f(x)-0 BUT my teacher said this--

Note that the equation has a degree of 2. f(x)=x^2-6x+5. How many turning points must an equation have that is of degree two? Is it two points--(1,0)and (5,0)

c. end behavior

I had for a large |x|(positiveor negative) f(x) increases as parabola upward to infinity BUT my teacher said not to change to an absolute value. Can someone help me with this one.

**please show work**

- calculus -
**Reiny**, Tuesday, December 18, 2012 at 9:51pmYour function is a simple parabola

All parabolas have one turning point, called the vertex

because of the + understood in front of

f(x) = +x^2 ....

the parabola opens upwards so it rises to infinity in both the 1st and 2nd quadrants.

If the function had been

f(x) = -x^2 .... , then it would have opened downwards. Your teacher was right to object to the |x| , that would always make the sign in front a +

Your zeros of x=1 and x=5 are correct, their multiplicity is 1

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