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Three points on the graph of the function 
f(x) are 
{(0,2),(1,4),(2,6)}. Which equation represents 
f(x) ?

f(x)=2*2^(x)

f(x)=2x+2

f(x)=(1)/(2)x-1

f(x)=x^(2)+2

Three points on the graph of the function f(x) f(x) are {(0,2),(1,4),(2,6)} \{(0,2),(1,4),(2,6)\} . Which equation represents f(x) f(x) ?\newlinef(x)=22x f(x)=2 \cdot 2^{x} \newlinef(x)=2x+2 f(x)=2 x+2 \newlinef(x)=12x1 f(x)=\frac{1}{2} x-1 \newlinef(x)=x2+2 f(x)=x^{2}+2

Full solution

Q. Three points on the graph of the function f(x) f(x) are {(0,2),(1,4),(2,6)} \{(0,2),(1,4),(2,6)\} . Which equation represents f(x) f(x) ?\newlinef(x)=22x f(x)=2 \cdot 2^{x} \newlinef(x)=2x+2 f(x)=2 x+2 \newlinef(x)=12x1 f(x)=\frac{1}{2} x-1 \newlinef(x)=x2+2 f(x)=x^{2}+2
  1. Test Function 0,20, 2: Let's test each given function with the points provided to see which one fits all three points.\newlineFirst, we will test the point 0,20, 2 with each function.
  2. Test Function (11, 44): Testing the first function f(x)=22xf(x) = 2 \cdot 2^{x} with the point (0,2)(0, 2):f(0)=220=21=2f(0) = 2 \cdot 2^{0} = 2 \cdot 1 = 2. This matches the y-value of the point (0,2)(0, 2).
  3. Test Function (22, 66): Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2):f(0)=20+2=0+2=2.f(0) = 2\cdot0 + 2 = 0 + 2 = 2.This also matches the y-value of the point (0,2)(0, 2).
  4. Test Function (22, 66): Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2):f(0)=20+2=0+2=2.f(0) = 2\cdot 0 + 2 = 0 + 2 = 2.This also matches the y-value of the point (0,2)(0, 2).Testing the third function f(x)=12x1f(x) = \frac{1}{2}x - 1 with the point (0,2)(0, 2):f(0)=1201=01=1.f(0) = \frac{1}{2}\cdot 0 - 1 = 0 - 1 = -1.This does not match the y-value of the point (0,2)(0, 2). We can eliminate this function.
  5. Test Function (22, 66): Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2):
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 2.
    This also matches the y-value of the point (0,2)(0, 2).Testing the third function f(x)=12x1f(x) = \frac{1}{2}x - 1 with the point (0,2)(0, 2):
    f(0)=1201=01=1f(0) = \frac{1}{2}\cdot0 - 1 = 0 - 1 = -1.
    This does not match the y-value of the point (0,2)(0, 2). We can eliminate this function.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2):
    (0,2)(0, 2)00.
    This matches the y-value of the point (0,2)(0, 2).
  6. Test Function (22, 66): Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2):
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 2.
    This also matches the y-value of the point (0,2)(0, 2).Testing the third function f(x)=(12)x1f(x) = (\frac{1}{2})x - 1 with the point (0,2)(0, 2):
    f(0)=(12)01=01=1f(0) = (\frac{1}{2})\cdot0 - 1 = 0 - 1 = -1.
    This does not match the y-value of the point (0,2)(0, 2). We can eliminate this function.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2):
    (0,2)(0, 2)00.
    This matches the y-value of the point (0,2)(0, 2).Now, let's test the remaining functions with the second point (0,2)(0, 2)22.
  7. Test Function (2,6)(2, 6): Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2):f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 2. This also matches the y-value of the point (0,2)(0, 2).Testing the third function f(x)=12x1f(x) = \frac{1}{2}x - 1 with the point (0,2)(0, 2):f(0)=1201=01=1f(0) = \frac{1}{2}\cdot0 - 1 = 0 - 1 = -1. This does not match the y-value of the point (0,2)(0, 2). We can eliminate this function.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2):f(x)=2x+2f(x) = 2x + 211. This matches the y-value of the point (0,2)(0, 2).Now, let's test the remaining functions with the second point f(x)=2x+2f(x) = 2x + 233.Testing the first function f(x)=2x+2f(x) = 2x + 244 with the point f(x)=2x+2f(x) = 2x + 233:f(x)=2x+2f(x) = 2x + 266. This matches the y-value of the point f(x)=2x+2f(x) = 2x + 233.
  8. Test Function (22, 66): Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2):f(0)=20+2=0+2=2.f(0) = 2\cdot 0 + 2 = 0 + 2 = 2.This also matches the y-value of the point (0,2)(0, 2).Testing the third function f(x)=12x1f(x) = \frac{1}{2}x - 1 with the point (0,2)(0, 2):f(0)=1201=01=1.f(0) = \frac{1}{2}\cdot 0 - 1 = 0 - 1 = -1.This does not match the y-value of the point (0,2)(0, 2). We can eliminate this function.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2):f(0)=02+2=0+2=2.f(0) = 0^2 + 2 = 0 + 2 = 2.This matches the y-value of the point (0,2)(0, 2).Now, let's test the remaining functions with the second point (1,4)(1, 4).Testing the first function (0,2)(0, 2)00 with the point (1,4)(1, 4):f(1)=221=22=4.f(1) = 2\cdot 2^{1} = 2\cdot 2 = 4.This matches the y-value of the point (1,4)(1, 4).Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (1,4)(1, 4):f(1)=21+2=2+2=4.f(1) = 2\cdot 1 + 2 = 2 + 2 = 4.This also matches the y-value of the point (1,4)(1, 4).
  9. Test Function (22, 66): Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2):
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 2.
    This also matches the y-value of the point (0,2)(0, 2).Testing the third function f(x)=12x1f(x) = \frac{1}{2}x - 1 with the point (0,2)(0, 2):
    f(0)=1201=01=1f(0) = \frac{1}{2}\cdot0 - 1 = 0 - 1 = -1.
    This does not match the y-value of the point (0,2)(0, 2). We can eliminate this function.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2):
    (0,2)(0, 2)00.
    This matches the y-value of the point (0,2)(0, 2).Now, let's test the remaining functions with the second point (0,2)(0, 2)22.Testing the first function (0,2)(0, 2)33 with the point (0,2)(0, 2)22:
    (0,2)(0, 2)55.
    This matches the y-value of the point (0,2)(0, 2)22.Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2)22:
    (0,2)(0, 2)99.
    This also matches the y-value of the point (0,2)(0, 2)22.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2)22:
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 233.
    This does not match the y-value of the point (0,2)(0, 2)22. We can eliminate this function.
  10. Test Function (22, 66): Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2):
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 2.
    This also matches the y-value of the point (0,2)(0, 2).Testing the third function f(x)=12x1f(x) = \frac{1}{2}x - 1 with the point (0,2)(0, 2):
    f(0)=1201=01=1f(0) = \frac{1}{2}\cdot0 - 1 = 0 - 1 = -1.
    This does not match the y-value of the point (0,2)(0, 2). We can eliminate this function.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2):
    (0,2)(0, 2)00.
    This matches the y-value of the point (0,2)(0, 2).Now, let's test the remaining functions with the second point (0,2)(0, 2)22.Testing the first function (0,2)(0, 2)33 with the point (0,2)(0, 2)22:
    (0,2)(0, 2)55.
    This matches the y-value of the point (0,2)(0, 2)22.Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2)22:
    (0,2)(0, 2)99.
    This also matches the y-value of the point (0,2)(0, 2)22.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2)22:
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 233.
    This does not match the y-value of the point (0,2)(0, 2)22. We can eliminate this function.Finally, let's test the remaining functions with the third point f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 255.
  11. Test Function (22, 66): Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2):
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 2.
    This also matches the y-value of the point (0,2)(0, 2).Testing the third function f(x)=12x1f(x) = \frac{1}{2}x - 1 with the point (0,2)(0, 2):
    f(0)=1201=01=1f(0) = \frac{1}{2}\cdot0 - 1 = 0 - 1 = -1.
    This does not match the y-value of the point (0,2)(0, 2). We can eliminate this function.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2):
    (0,2)(0, 2)00.
    This matches the y-value of the point (0,2)(0, 2).Now, let's test the remaining functions with the second point (0,2)(0, 2)22.Testing the first function (0,2)(0, 2)33 with the point (0,2)(0, 2)22:
    (0,2)(0, 2)55.
    This matches the y-value of the point (0,2)(0, 2)22.Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2)22:
    (0,2)(0, 2)99.
    This also matches the y-value of the point (0,2)(0, 2)22.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2)22:
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 233.
    This does not match the y-value of the point (0,2)(0, 2)22. We can eliminate this function.Finally, let's test the remaining functions with the third point f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 255.Testing the first function (0,2)(0, 2)33 with the point f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 255:
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 288.
    This does not match the y-value of the point f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 255. We can eliminate this function.
  12. Test Function (22, 66): Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2):
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 2.
    This also matches the y-value of the point (0,2)(0, 2).Testing the third function f(x)=12x1f(x) = \frac{1}{2}x - 1 with the point (0,2)(0, 2):
    f(0)=1201=01=1f(0) = \frac{1}{2}\cdot0 - 1 = 0 - 1 = -1.
    This does not match the y-value of the point (0,2)(0, 2). We can eliminate this function.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2):
    (0,2)(0, 2)00.
    This matches the y-value of the point (0,2)(0, 2).Now, let's test the remaining functions with the second point (0,2)(0, 2)22.Testing the first function (0,2)(0, 2)33 with the point (0,2)(0, 2)22:
    (0,2)(0, 2)55.
    This matches the y-value of the point (0,2)(0, 2)22.Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point (0,2)(0, 2)22:
    (0,2)(0, 2)99.
    This also matches the y-value of the point (0,2)(0, 2)22.Testing the fourth function f(x)=x2+2f(x) = x^2 + 2 with the point (0,2)(0, 2)22:
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 233.
    This does not match the y-value of the point (0,2)(0, 2)22. We can eliminate this function.Finally, let's test the remaining functions with the third point f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 255.Testing the first function (0,2)(0, 2)33 with the point f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 255:
    f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 288.
    This does not match the y-value of the point f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 255. We can eliminate this function.Testing the second function f(x)=2x+2f(x) = 2x + 2 with the point f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 255:
    (0,2)(0, 2)22.
    This matches the y-value of the point f(0)=20+2=0+2=2f(0) = 2\cdot0 + 2 = 0 + 2 = 255. Since this is the only function that fits all three points, it must be the correct representation of (0,2)(0, 2)44.

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