look at the following points to see if they represent a linear, quadratic, or exponential function, show your work, & explain your reasoning. (-1, -3), (0, -1), (1, 5), (2, 15)

Question

look at the following points to see if they represent a linear, quadratic, or exponential function, show your work, & explain your reasoning. (-1, -3), (0, -1), (1, 5), (2, 15)

Answer 1

Check for linear

y = mx+b

-3 = -m+b 1st Point

-1 = b

m = 4 2nd Point

y = 4x-1

Test 3rd point

5 ?= 4 -1 No

Check for quadratic

y = ax²+bx+c

-3 = a - b + c

-1 = c

5 = a+b-1

a+b = 6

a-b = -2

2a = 4 --> a = 2 b=4

y = 2x²+4x-1

15 ?= 2(4)+8-1 = 16-1 = 15 Yes

Exponential

y = ae^bx+c

-3 = ae^-b+c eq1

-1 = a +c --> c = -1-a eq2

5 = ae^b+c = ae^b-1-a eq3

Substituting c into eq1 and eq3

a-2 = ae^-b

a+6 = ae^b

Taking logs

ln(a-2) = ln(a) - b

ln(a+6) = ln(a) + b

Adding to remove b

ln(a-2)+ln(a+6) = 2ln(a) = ln(a²)

ln((a-2)(a+6)) = ln(a²)

a²+4a-12 = a²

a = 3

b = ln((a+6)/a)) = ln(9/3) = ln(3)

c = -4

y = 3e^ln(3)x-4= 3(3)^x-4= 3^x+1-4

Checking points

-1 = 3⁰-4 = -1 yes

-1 = 3¹-4 = -1 yes

5 = 3²-4 = 9-4 = 5 yes

15 = 3³ -4 = 27 -4 = 23 No

Points only fit a qudratic

Answer 2

Plot them, they're not all on a straight line. It's not linear, although it's relatively close to a straight line.

It's either exponential or quadratic.

y=a(b)^x + c won't work if you plug in the 4 points. You can get it to fit 3 points but not all 4. y=3^(x+1) -4 works for 3 points but not (2,15)

y=a(x-h)^2 + k works for a=2, h=-1, k=-3,

y=2(x+1)-3. Plug in each of the 4 points, they all work. A quadratic function fits all 4 points

a quadratic best fits the 4 points