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Ivenika [448]
2 years ago
12

Systems of equations with graphing

Mathematics
1 answer:
NARA [144]2 years ago
4 0
X is 2 + y/2 & y is 0
you can solve for x in the first equation
once you solve for x
you plug your answer to the equation again and solve for y
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Hunter-Best [27]

(a) If

<em>f(x)</em> = <em>a</em>₀ + <em>a</em>₁ <em>x </em>+<em> a</em>₂ <em>x</em> ² + <em>a</em>₃ <em>x</em> ³

then from the given conditions we get the system of equations,

<em>f</em> (-1) = <em>a</em>₀ - <em>a</em>₁<em> </em>+<em> a</em>₂ - <em>a</em>₃ = -1

<em>f</em> (1) = <em>a</em>₀ + <em>a</em>₁<em> </em>+<em> a</em>₂ + <em>a</em>₃ = 2

<em>f</em> (2) = <em>a</em>₀ + 2<em>a</em>₁<em> </em>+ 4<em>a</em>₂ + 8<em>a</em>₃ = 1

<em>f</em> (3) = <em>a</em>₀ + 3<em>a</em>₁<em> </em>+<em> </em>9<em>a</em>₂ + 27<em>x</em> ³ = 5

(b) Similarly, if

<em>f(x)</em> = <em>a</em>₀ + <em>a</em>₁ <em>x </em>+<em> a</em>₂ <em>x</em> ² + <em>a</em>₃ <em>x</em> ³

then

<em>f'(x)</em> = <em>a</em>₁<em> </em>+<em> </em>2<em>a</em>₂ <em>x</em> + 3<em>a</em>₃ <em>x</em> ²

so that the given conditions yield the system,

<em>f</em> (1) = <em>a</em>₀ + <em>a</em>₁<em> </em>+<em> a</em>₂ + <em>a</em>₃ = 0

<em>f'</em> (1) = <em>a</em>₁<em> </em>+<em> </em>2<em>a</em>₂ + 3<em>a</em>₃ = 2

<em>f</em> (2) = <em>a</em>₀ + 2<em>a</em>₁<em> </em>+<em> </em>4<em>a</em>₂ + 27<em>a</em>₃ = 3

<em>f'</em> (2) = <em>a</em>₁<em> </em>+<em> </em>4<em>a</em>₂ + 12<em>a</em>₃ = -1

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