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How to find the a in the equation y=ax^3 + d given the two points (0,10), (2,20)

Sagot :

semsee45

Answer:

[tex]y=\dfrac{5}{4}x^3+10[/tex]

Step-by-step explanation:

Given information:

  • [tex]y=ax^3+d[/tex]
  • (0, 10)
  • (2, 20)

Create two equations by substituting the given points into the given equation:

Equation 1:  point (0, 10)

[tex]\implies a(0)^3+d=10[/tex]

[tex]\implies 0+d=10[/tex]

[tex]\implies d=10[/tex]

Equation 2:  point (2, 20)

[tex]\implies a(2)^3+d=20[/tex]

[tex]\implies 8a+d=20[/tex]

Substitute Equation 1 into Equation 2 and solve for a:

[tex]\implies 8a+d=20[/tex]

[tex]\implies 8a+10=20[/tex]

[tex]\implies 8a+10-10=20-10[/tex]

[tex]\implies 8a=10[/tex]

[tex]\implies \dfrac{8a}{8}=\dfrac{10}{8}[/tex]

[tex]\implies a=\dfrac{10}{8}[/tex]

[tex]\implies a=\dfrac{5}{4}[/tex]

Finally, substitute the found values of a and d into the original formula:

[tex]\implies y=\dfrac{5}{4}x^3+10[/tex]

Check by substituting the x-values of the two given points into the found equation:

[tex]x=0 \implies y=\dfrac{5}{4}(0)^3+10=10 \leftarrow \textsf{correct}[/tex]

[tex]x=2 \implies y=\dfrac{5}{4}(2)^3+10=20 \leftarrow \textsf{correct}[/tex]
  • y=ax³+d

Put(0,10)

  • 10=a(0)³+d
  • d=10

Now

Put again (2,20) this time

  • 20=2³a+10
  • 10=8a
  • a=10/8
  • a=5/4
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