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The function f(x) = –x3 – 7x2 – 7x + 15 has zeros located at –5, –3, 1. Verify the zeros of f(x) and explain how you verified them. Describe the end behavior of the function.

Please help! I need this done today & I'm not sure how to go about it. (Detailed explanation please)

User Anand S
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1 Answer

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Answer/Step-by-step explanation:

Zeros are x-intercepts. (Solutions are zeros are x-intercepts!)

If you are allowed to use technology, such as a graphing calculator or an online graphing tool, input the function and look for the x-intercepts. Graphing calculators have menus to find them for you. It is a legit way to "verify the zeros" Also, since this is a third degree function (highest exponent is a 3rd power) but the leading coefficient is negative, the end behavior acts the same as:

y = -x^3

a much simpler version. Actually, even simpler:

y = -x

All of these, go up to the left and down to the right. Or some teachers/classes/books/programs want you to say:

as x goes to negative infinity, y goes to infinity (up to the left) and as x goes to infinity, y goes to negative infinity (down to the right)

SO, if you cannot use technology, put the given x-intercept into the function. And check that it gives you a value of 0. Like this:

f(x) =

-x^3-7x^2-7x+15

check by substituting -5 for x

f(-5) =

-(-5)^3-7(-5)^2-7(-5)+15

= 125 -7(25) +35+15

= 125- 175 + 35 + 15

= 0

When you use x =-5 you get y = 0. This verifies that -5 is a zero.

Check -3.

f(-3)=-(-3)^3-7(-3)^2-7(-3)+15

= 27 -7(9) +21 +15

= 27 - 63 +21 +15

= 0

Verified!

Lastly, check 1:

f(1)=-(1)^3-7(1)^2-7(1)+15

= -1 - 7 - 7 + 15

= 0

Verified!

User Timothy Zorn
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