- #1

- 15

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a) (25, 43)

b) (28, 46)

c) (32, 47)

d) (35, 45)

would i use y=mx+b i think thats a straight line

- Thread starter lucifer_x
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- #1

- 15

- 0

a) (25, 43)

b) (28, 46)

c) (32, 47)

d) (35, 45)

would i use y=mx+b i think thats a straight line

- #2

gabbagabbahey

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[tex]y(x)=Ax^3+Bx^2+Cx+D[/tex]

Where you can determine the constants [itex]A[/itex], [itex]B[/itex], [itex]C[/itex] and [itex]D[/itex] by plugging in the given points and solving the resulting system of 4 equations.

- #3

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[tex]y(x)=Ax^3+Bx^2+Cx+D[/tex]

cause there are 2 points and only 1 A

- #4

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If you are studying

- #5

gabbagabbahey

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Unfortunately the points don't all lie on the same quadratic, so a cubic is necessary.my guess is that you only need [tex]y(x) = Ax^2 + Bx + C[/tex].

- #6

gabbagabbahey

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I thought a), b) c) and d) were each points on the curve in the form [itex](x_1,y_1)=(25,43)[/itex] etc.....is this assumption incorrect?

[tex]y(x)=Ax^3+Bx^2+Cx+D[/tex]

cause there are 2 points and only 1 A

- #7

- 15

- 0

so like

[itex](x_1,y_1)=(25,43)[/itex]

is what i was thinking

but thats only one point

- #8

gabbagabbahey

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[tex]43=A(25)^3+B(25)^2+C(25)+D[/tex]

And so using all 4 points you get 4 equations and 4 unknowns ([itex]A[/itex], [itex]B[/itex], [itex]C[/itex], and [itex]D[/itex]) allowing you to solve for them.

- #9

- 15

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so the almost final will look like this

[tex]43=A(15625)+B(625)+C(25)+D[/tex]

then you divide them ??

- #10

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Equation 1: [tex]y_1(x)=Ax^3_1+Bx^2_1+Cx_1+D[/tex]

Equation 2: [tex]y_2(x)=Ax^3_2+Bx^2_2+Cx_2+D[/tex]

Equation 3: [tex]y_3(x)=Ax^3_3+Bx^2_3+Cx_3+D[/tex]

Equation 4: [tex]y_4(x)=Ax^3_4+Bx^2_4+Cx_4+D[/tex]

Plug in your coordinates for your (x,y) values above and then solve the system.

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