Find a cubic polynomial that goes through points (4, – 22) and (3, - 26) and has tangents with slopes respectively 11 and — 2 there. Check your work with a graphing utility.
f() =

Let f(x) = ax ³ + bx ² + cx + d.

The graph of f(x) passes through (4, -22) and (3, -26), which means f (4) = -22 and f (3) = -26, so that

64a + 16b + 4c + d = -22

27a + 9b + 3c + d = -26

When the question says it has tangents at some point, I take that to mean the slope of the tangent line at that point is the given number. So f ' (4) = 11 and f ' (3) = -2. We have

f '(x) = 3ax ²+ 2bx + c

so that

48a + 8b + c = 11

27a + 6b + c = -2

Solve the system:

• Eliminate d :

(64a + 16b + 4c + d) - (27a + 9b + 3c + d) = -22 - (-26)

→   37a + 7b + c = 4

• Eliminate c :

(48a + 8b + c) - (27a + 6b + c) = 11 - (-2)

→   21a + 2b = 13

(48a + 8b + c) - (37a + 7b + c) = 11 - 4

→   11a + b = 7

• Eliminate b, then solve for a and the other variables:

(21a + 2b) - 2 (11a + b) = 13 - 2 (7)

-a = -1

a = 1   →   b = -4   →   c = -5   →   d = -2

Then

f(x) = x ³ - 4x ² - 5x - 2

the answe i =−n+5

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13