Find a polynomial of the form f(x)=ax^3+bx^2+cx+d such that f(0)=−5, f(−2)=6, f(4)=5, and f(−5)=−3. f(x) = ?

Answer 1

You are given 4 data points:

f(0) = -5 means the point (0, -5) is on the curve.

f(-2) = 6 means the point (-2, 6) is on the curve.

f(4) = 5 means the point (4, 5) is on the curve.

f(-5) = -3 means the point (-5, -3) is on the curve.

So, you can plug in each (x, y) to get 4 equations in 4 unknowns:

Equation #1: -5 = a(0)³ + b(0)² + c(0) + d or -5 = d

Based on that, you can plug in d = -5 into the other equations giving you 3 equations in 3 unknowns:

Equation #2: 6 = a(-2)³ + b(-2)² + c(-2) - 5 or -8a + 4b - 2c = 11

Equation #3: 5 = a(4)³ + b(4)² + c(4) - 5 or 64a + 16b + 4c = 10 or 32a + 8b + 2c = 5

Equation #4: -3 = a(-5)³ + b(-5)² + c(-5) - 5 or -125a + 25b - 5c = 2

Adding Equations 2 & 3:

-8a + 4b - 2c = 11

32a + 8b + 2c = 5

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24a + 12b = 16

or 6a + 3b = 4

So Equation #5 is 6a + 3b = 4

Multiplying Equation #4 by 2 and multiplying Equation #3 by 5 and adding them:

-250a + 50b - 10c = 4

160a + 40b + 10c = 25

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-90a + 90b = 29

So Equation #6 is -90a + 90b = 29

Multiplying Equation #5 by -30 and adding it to Equation #6:

-180a - 90b -120

-90a + 90b = 29

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-270a = -91

a = 91/270

Plugging a = 91/270 into Equation #5:

6(91/270) + 3b = 4

91/45 + 3b = 4

3b = 180/45 - 91/45

3b = 89/45

b = 89/135

Plugging in b = 89/135 and a = 91/270 into Equation #2:

-8(91/270) + 4(89/135) - 2c = 11

-4(91/135) + 4(89/135) - 11 = 2c

-364/135 + 356/135 - 1485/135 = 2c

-1493/135 = 2c

c = -1493/270

Plug the values into the function f(x) = ax³ + bx² + cx + d

Answer 2

f(x) = ax^3 + bx^2 + cx + d such that f(0) = −5, f(−2) = 6, f(4) = 5, and f(−5) = −3. f(x) = ?

f(x) = ax^3 + bx^2 + cx + d

f(0) = −5 ==> f(0) = a(0)^3 + b(0)^2 + c(0) + d = −5 ==> 0a + 0b + 0c + d = −5 ==> d = −5

f(−2) = 6 ==> f(−2) = a(−2)^3 + b(−2)^2 + c(−2) + d = 6 ==> -8a + 4b - 2c − 5 = 6 ==> -8a + 4b - 2c = 11

f(4) = 5 ==> f(4) = a(4)^3 + b(4)^2 + c(4) + d = 5 ==> 64a + 16b + 4c − 5 = 5 ==> 32a + 8b + 2c = 5

f(−5) = −3 ==> f(−5) = a(−5)^3 + b(−5)^2 + c(−5) + d = −3 ==> -125a + 25b - 5c − 5 = −3 ==> -125a + 25b - 5c = 2

1. d = −5
2. -8a + 4b - 2c = 11
3. 32a + 8b + 2c = 5
4. -125a + 25b - 5c = 2

In TI-84 Plus, create 3X4 Matrix A and use rref(A) to solve Matrix A. The result is:

a = 91/270

b = 89/135

c = -1493/270

d = −5

f(x) = ax^3 + bx^2 + cx + d

f(x) = (91/270)x^3 + (89/135)x^2 - (1493/270)x − 5

f(x) = (91/270)x³ + (89/135)x² - (1493/270)x − 5 (Final Answer)