The value of the expression when t = 32 degrees is
degrees.

The zeros of

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

are

x=0,x=-2,x=-5x=0,x=−2,x=−5

EXPLANATION

The given function is

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

To find the zeros of the above function, we just have to equate the function to zero and solve for x.

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

This implies that,

x = 0 \: \: or \: x + 2 = 0 \:or \: x + 5 = 0x=0orx+2=0orx+5=0

x = 0 \: \: or \: x = - 2\:or \: x = - 5x=0orx=−2orx=−5

To graph the above function, we need to consider the multiplicity.

We can see that the multiplicity of the roots are odd. This means that, the graph crosses the x-axis at each x-intercept.

We also need to consider the position of the graph on the following intervals,

x < - 5x<−5

When

x = - 10x=−10

y = - 10( - 8)( - 5) \: < \: 0y=−10(−8)(−5)<0

The graph is below the x-axis.

- 5 \: < \: x \: < \: - 2−5<x<−2

When

x = - 3x=−3

y = - 3(2)( - 1) \: > 0y=−3(2)(−1)>0

The graph is above the x-axis.

- 2 \: < \: x < \: 0−2<x<0

when

x = - 1x=−1

y = -1(1)(4) \: < \: 0y=−1(1)(4)<0

The graph is below the x-axis.

Finally the interval,

x \: > \: 0x>0

when

x = 1x=1

y = 1(3)(6) \: > \: 0y=1(3)(6)>0

The graph is above the x-axis.

We can now use the above information to sketch graph as shown in the diagram above