Y=f(x)=x^2+x
find the slope of the secant line joining (-3,f(-3)) and (0,f(0))

-2

The problem:

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

find the slope of the secant line joining (-3,f(-3)) and (0,f(0))

Step-by-step explanation:

We need to first find the values that correspond to f(-3) and f(0).

f(-3) can be found by replacing the x's in f(x)=x^2+x with (-3):

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

f(-3)=9+-3

f(-3)=6

f(0) can be found by replacing the x's in f(x)=x^2+x with (0):

f(0)=(0)^2+(0)

f(0)=0+0

f(0)=0

So we want to find the slope of the line going through the points:

(-3,6) and (0,0)

Line them up-doesn't matter which point goes on top:

(-3,6)

(0,0)

Now subtract!

-3 , 6

The rise goes over the run so the slope is 6/-3 which simplifies to -2.

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

answer: 1 and 3

step-by-step explanation: