Answer:
(c) For p = 15, leaves a remainder of -2 when divided by (x-3).
Step-by-step explanation:
Here, The dividend expression is = E(x)
The Divisor = (x-3)
Remainder = -2
Now, by <u>REMAINDER THEOREM</u>:
Dividend = (Divisor x Quotient) + Remainder
If ( x -3 ) divides the given polynomial with a remainder -2.
⇒ x = 3 is a solution of given polynomial E(x) - (-2) =
= S(x)
Now, S(3) = 0
⇒
or, p =1 5
Hence, for p = 15, leaves a remainder of -2 when divided by (x-3).
well, about A and D, I just plugged the values on the slope formula of
for A the values are 8.01 and 8.0, so indeed those "slopes" are close.
for D the values are -2.25 and 8.0, so no dice on that one.
for B, let's check the y-intercept for g(x), by setting x = 0, we end up g(0) = 0³-4(0)+1, which gives us g(0) = 1.
checking L(x) y-intercept, well, L(x) is in slope-intercept form, thus the +1 sticking out on the far right is the y-intercept, so, dice.
for C, well, the slope if L(x) is 8, since it's in slope-intercept form, the derivative of g(x) is g'(x) = 3x² - 4, and thus g'(0) = -4, so no dice.
for E, do they intercept at (2,1)? well, come on now, L(x) is a tangent line to g(x), so that's a must for a tangent.
for F, we know the slope of the line L(x) is 8, is g'(2) = 8? let's check
recall that g'(x) = 3x² - 4, so g'(2) = 3(2)² - 4, meaning g'(2) = 8, so, dice.