which of the following exponential regression equations best fits the data shown below?\n| x | -4 | -3 | -2…

Explanation:

Step1: Use x=0 to test

For an exponential function ( y = a \cdot b^x ), when ( x = 0 ), ( y=a \cdot b^0=a ). From the data, when ( x = 0 ), ( y = 5 ). So we check the 'a' value (the coefficient) of each option:

• Option A: ( a = 6.37 ) (close to 5 but not exact yet, check other x)
• Option B: ( a = 2.08 ) (too small)
• Option C: ( a = 3.53 ) (too small)
• Option D: ( a = 7.13 ) (too big)

Step2: Test x=1

When ( x = 1 ), ( y = 16 ). Let's check each option:

• Option A: ( y=6.37 \cdot 5.02^1\approx6.37\times5.02\approx31.98 ) (close to 16? No, but wait, maybe miscalculation. Wait, no, let's recalculate. Wait, 6.375.02: 65=30, 60.02=0.12, 0.375=1.85, 0.37*0.02=0.0074, total ≈30+0.12+1.85+0.0074≈31.9774.
• Option B: ( y=2.08 \cdot 2.61^1 = 2.08\times2.61\approx5.43 ) (too small)
• Option C: ( y=3.53 \cdot 4.37^1 = 3.53\times4.37\approx15.43 ) (very close to 16)
• Option D: ( y=7.13 \cdot 6.09^1 = 7.13\times6.09\approx43.42 ) (too big)

Wait, maybe x=0 was a start, but let's check x= - 1. When ( x=-1 ), ( y = 0.75 ). For an exponential function ( y=a\cdot b^x ), when ( x=-1 ), ( y=\frac{a}{b} ).

• Option A: ( \frac{6.37}{5.02}\approx1.27 ) (not 0.75)
• Option B: ( \frac{2.08}{2.61}\approx0.797 ) (close to 0.75)
• Option C: ( \frac{3.53}{4.37}\approx0.808 ) (close to 0.75)
• Option D: ( \frac{7.13}{6.09}\approx1.17 ) (no)

Wait, maybe better to use x=2. When ( x = 2 ), ( y = 65 ).

• Option A: ( y=6.37 \cdot 5.02^2\approx6.37\times25.2004\approx160.53 ) (too big)
• Option B: ( y=2.08 \cdot 2.61^2 = 2.08\times6.8121\approx14.17 ) (too small)
• Option C: ( y=3.53 \cdot 4.37^2 = 3.53\times(4.37\times4.37)=3.53\times19.0969\approx67.41 ) (very close to 65)
• Option D: ( y=7.13 \cdot 6.09^2 = 7.13\times37.0881\approx264.44 ) (too big)

x=3: ( y = 276 )

• Option C: ( y=3.53 \cdot 4.37^3 ). First, ( 4.37^2 = 19.0969 ), ( 4.37^3 = 4.37\times19.0969\approx83.45 ). Then ( 3.53\times83.45\approx294.58 ) (close to 276, slight difference, but others are worse)
• Option A: way too big, Option B: too small, Option D: too big.

Wait, maybe I made a mistake with x=0. Wait the data at x=0 is y=5. Let's check Option C: a=3.53, but when x=0, y=3.53, but data is 5. Hmm, maybe my initial step was wrong. Wait, maybe the exponential regression is not perfect, but we need to see the trend.

Wait, let's calculate the ratio between consecutive y-values (for x>0) to see the base 'b'.

From x=0 to x=1: 16/5 = 3.2

x=1 to x=2: 65/16 ≈4.0625

x=2 to x=3: 276/65≈4.246

x=3 to x=4: 1251/276≈4.532

So the base 'b' is around 4 - 5. Let's check the bases:

Option A: b=5.02

Option B: b=2.61

Option C: b=4.37

Option D: b=6.09

Now, the coefficient 'a' when x=0 is y=a, so a should be around 5. Wait Option A: a=6.37, close to 5. Let's check x= - 4: y=0.01.

Option A: ( y=6.37 \cdot 5.02^{-4}=\frac{6.37}{5.02^4} ). ( 5.02^2=25.2004 ), ( 5.02^4=(25.2004)^2≈635.06 ). So ( 6.37/635.06≈0.01003 ), which is very close to 0.01 (the y-value at x=-4).

Let's check x=-3: ( y=6.37 \cdot 5.02^{-3}=\frac{6.37}{5.02^3} ). ( 5.02^3=5.02\times25.2004≈126.506 ). ( 6.37/126.506≈0.0503 ), but the data is 0.04. Close.

x=-2: ( y=6.37 \cdot 5.02^{-2}=\frac{6.37}{25.2004}≈0.2528 ), but data is 0.16. Hmm, not perfect. Wait x=-1: ( y=6.37 \cdot 5.02^{-1}=\frac{6.37}{5.02}≈1.27 ), but data is 0.75.

Wait Option C: x=-4: ( y=3.53 \cdot 4.37^{-4}=\frac{3.53}{4.37^4} ). ( 4.37^2=19.0969 ), ( 4.37^4=(19.0969)^2≈364.69 ). ( 3.53/364.69≈0.00968 ), close to 0.01.

x=-3: ( y=3.53 \cdot 4.37^{-3}=\frac{3.53}{4.37^3} ). ( 4.37^3=4.37\times19.0969≈83.45 ). ( 3.53/83.45≈0.0423 ), close to 0.04.

x=-2: ( y=3.53 \cdot 4.37^{-2}=\frac{3.53}{19.0969}≈0.1849 ), data is 0.16. Close.

x=-1: ( y=3.53 \cdot 4.37^{-1}=\frac{3.53}{4.37}≈0.8078 ), data is 0.75. Close.

x=0: 3.53, data is 5. Hmm, difference here.

x=1: 3.53*4.37≈15.43, data is 16. Close.

x=2: 3.53*(4.37)^2≈3.53*19.0969≈67.41, data is 65. Close.

x=3: 3.53*(4.37)^3≈3.53*83.45≈294.58, data is 276. Close.

x=4: 3.53*(4.37)^4≈3.53*364.69≈1287, data is 1251. Very close.

Wait, Option A at x=4: 6.37*(5.02)^4≈6.37*635.06≈4045, which is way bigger than 1251. So Option A is wrong at x=4.

Option C at x=4: ~1287, close to 1251.

Option D at x=4: 7.13*(6.09)^4. 6.09^2=37.0881, 6.09^4=(37.0881)^2≈1375.5, 7.13*1375.5≈9807, way too big.

Option B at x=4: 2.08*(2.61)^4. 2.61^2=6.8121, 2.61^4=46.39, 2.08*46.39≈96.5, way too small.

So the best fit is Option C? Wait no, wait when x=0, Option A has a=6.37, y=6.37, but data is 5. Option C has a=3.53, y=3.53, data is 5. But when x=4, Option C is 1287 vs 1251, Option A is 4045 vs 1251. So Option C is better. Wait but earlier when I checked x=1, Option C gave 15.43 vs 16, which is close. Option A gave 31.98 vs 16, which is way off. So the correct answer should be Option C? Wait no, wait let's recalculate Option A at x=1: 6.375.02=31.97, which is not 16. Option C: 3.534.37=15.43, close to 16. x=2: 3.53*(4.37)^2=3.5319.0969≈67.41, data is 65. Close. x=3: 3.53(4.37)^3=3.5383.45≈294.58, data is 276. Close. x=4: 3.53(4.37)^4≈3.53*364.69≈1287, data is 1251. Close. And for negative x:

x=-4: 3.53*(4.37)^-4≈3.53/364.69≈0.00968≈0.01 (data is 0.01)

x=-3: 3.53*(4.37)^-3≈3.53/83.45≈0.0423≈0.04 (data is 0.04)

x=-2: 3.53*(4.37)^-2≈3.53/19.0969≈0.1849≈0.16 (close)

x=-1: 3.53*(4.37)^-1≈3.53/4.37≈0.8078≈0.75 (close)

x=0: 3.53≈5? No, that's a problem. Wait the data at x=0 is 5, but Option C gives 3.53. Option A gives 6.37. Maybe the exponential regression is done by a calculator, and the best fit is Option A? Wait no, let's use a calculator approach. Let's take the natural log of y to linearize.

For exponential regression ( y = a b^x ), take ln(y) = ln(a) + x ln(b). Let's create a table of x and ln(y):

x: -4, -3, -2, -1, 0, 1, 2, 3, 4

ln(y): ln(0.01)≈-4.605, ln(0.04)≈-3.219, ln(0.16)≈-1.833, ln(0.75)≈-0.288, ln(5)≈1.609, ln(16)≈2.773, ln(65)≈4.174, ln(276)≈5.623, ln(1251)≈7.133

Now, perform linear regression on x and ln(y). Let's calculate the slope (m = ln(b)) and intercept (c = ln(a)).

First, calculate the mean of x (x̄) and mean of ln(y) (ȳ).

x values: -4, -3, -2, -1, 0, 1, 2, 3, 4. Sum of x: (-4)+(-3)+(-2)+(-1)+0+1+2+3+4 = 0. So x̄ = 0/9 = 0.

ln(y) values: -4.605, -3.219, -1.833, -0.288, 1.609, 2.773, 4.174, 5.623, 7.133. Sum of ln(y): Let's add them:

-4.605 -3.219 = -7.824; -7.824 -1.833 = -9.657; -9.657 -0.288 = -9.945; -9.945 +1.609 = -8.336; -8.336 +2.773 = -5.563; -5.563 +4.174 = -1.389; -1.389 +5.623 = 4.234; 4.234 +7.133 = 11.367. ȳ = 11.367 / 9 ≈1.263.

Now, slope m = (Σ(xi - x̄)(ln(yi) - ȳ)) / (Σ(xi - x̄)^2). Since x̄=0, this simplifies to (Σ(xi ln(yi))) / (Σ(xi^2)).

Calculate Σ(xi ln(yi)):

x=-4: -4*(-4.605)=18.42

x=-3: -3*(-3.219)=9.657

x=-2: -2*(-1.833)=3.666

x=-1: -1*(-0.288)=0.288

x=0: 0*1.609=0

x=1: 1*2.773=2.773

x=2: 2*4.174=8.348

x=3: 3*5.623=16.869

x=4: 4*7.133=28.532

Sum these: 18.42 +9.657=28.077; +3.666=31.743; +0.288=32.031; +0=32.031; +2.773=34.804; +8.348=43.152; +16.869=60.021; +28.532=88.553.

Σ(xi^2): (-4)^2 +