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algebra
what are the roots, and y intercept of g (x)?

f (x)=(x-2)(x+3)(x-4)(x+1)
g (x)= -f (x+3)​

User Bumperbox
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1 Answer

6 votes

Answers:

Roots of g(x) are: -1, -6, 1, -4

Y intercept of g(x) is: 24

==============================================

Step-by-step explanation:

Part 1) Finding the roots of g(x)

To do this, we need the roots of f(x). Plug in f(x) = 0 and solve for x using the zero product property.

f(x) = (x-2)(x+3)(x-4)(x+1)

0 = (x-2)(x+3)(x-4)(x+1)

(x-2)(x+3)(x-4)(x+1) = 0

x-2 = 0 or x+3 = 0 or x-4 = 0 or x+1 = 0

x = 2 or x = -3 or x = 4 or x = -1

Those four x values are the roots of f(x). Plugging any of those values into f(x) leads to f(x) = 0.

Now onto g(x). We see that g(x) = -f(x+3). The x+3 indicates the xy axis has been shifted by 3 units to the right. This gives the illusion the f(x) curve is shifted 3 units to the left. This will move each root 3 units to the left as well.

  • The root x = 2 on f(x) moves to x = -1 on g(x). We subtract 3 from the first x value to get the second x value.
  • The root x = -3 on f(x) moves to x = -6 on g(x). Same idea as above.
  • The root x = 4 on f(x) moves to x = 1 on g(x).
  • The root x = -1 on f(x) moves to x = -4 on g(x).

The four roots of g(x) are: -1, -6, 1, -4

------------------------------------------------

Part 2) The y intercept of g(x)

Let's plug x = 0 into g(x) to see what happens

g(x) = -f(x+3)

g(0) = -f(0+3) ... replace every x with 0

g(0) = -f(3)

We don't know f(3) yet, but we can plug x = 3 into f(x) to find out

f(x) = (x-2)(x+3)(x-4)(x+1)

f(3) = (3-2)(3+3)(3-4)(3+1)

f(3) = (1)(6)(-1)(4)

f(3) = -24

Therefore,

g(0) = -f(3)

g(0) = -(-24)

g(0) = 24

The y intercept of g(x) is 24.

User Dencey
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