2x(x+5)^2(x-3)

Question

For each function (a) apply the leading-term test (b) determine the zero's and state the multiplicity of any repeated zero's (c) find a few additional points and then (d) graph the function 

Answer 1

Before doing anything else, we need to simplify the equation y=2x(x+5)²(x-3)

 

y=2x(x+5)(x+5)(x-3)

y=2x(x² +10x+25)(x-3)

y=2x(x³ -3x² +10x² -30x+25x -75)

y=2x(x³ -7x² -5x-75)

y=2x⁴ -14x³ -10x²-150x

 

now we are ready to start. 

a)for the leading term test we look at the term with the highest degree of x. In this case, it is 2x⁴. we are looking for the end behavior of the graph. Basically, when x gets really big (positive or negative) the x⁴ will be so much bigger it will "overpower" the other terms and control the shape of the graph. When x is really large, say 1,000,000 what does the leading term do (not specific just want to know if it is positive on negative). In this case it is  a really big positive number so our graph goes to positive infinity to the right (right because 1,000,000 is positive). For the left side we plug in a large negative, like -1,000,000. The negative is raise to an even power so it is also positive.

 

b)To find zeros it is easiest to look at the equation in factored form (bolded above). To find the zeros we set y=0

 

0=2x(x+5)(x+5)(x-3)

 

using zero product property we set we set each factor equal to zero

0=2x

0=x+5

0=x+5

0=x-3

 

the solution to each of these four equations gives us a zero. Since x=-5 occurs twice it has multiplicity 2. 

 

c and d) choose a few values for x. 0,1,-1 are always good choices and connect the dots remembering the end behavior we found in part a and the zeros (that is where graph touches x-axis).

 

*also zeros with odd multiplicity (like one) will cross the axis, zeros with even multiplicity will just touch and "bounce off"